Simultaneous Linear Equations

Given,

Equations : x + 2y = 1, 3x - y = 17

⇒ x + 2y = 1

⇒ x = 1 - 2y     ....(1)

Substituting value of x from equation (1) in 3x - y = 17, we get :

⇒ 3(1 - 2y) - y = 17

⇒ 3 - 6y - y = 17

⇒ -7y = 17 - 3

⇒ -7y = 14

⇒ y = $-\frac{14}{7}$ = -2.

Substituting value of y in equation (1), we get :

⇒ x = 1 - 2y

⇒ x = 1 - 2(-2)

⇒ x = 1 + 4

⇒ x = 5.

Hence, x = 5, y = -2.

Given,

Equations : 5x + 4y = 4, x - 12y = 20

⇒ x - 12y = 20

⇒ x = 20 + 12y     ....(1)

Substituting value of x from equation (1) in 5x + 4y = 4, we get :

⇒ 5(20 + 12y) + 4y = 4

⇒ 100 + 60y + 4y = 4

⇒ 100 + 64y = 4

⇒ 64y = 4 - 100

⇒ 64y = -96

⇒ y = $-\frac{96}{64} = -\frac{3}{2}$.

Substituting value of y in equation (1), we get :

⇒ x = 20 + $12 \times -\Big(\frac{3}{2}\Big)$

⇒ x = 20 + 6(-3)

⇒ x = 20 - 18

⇒ x = 2.

Hence, x = 2, y = $-\frac{3}{2}$.

Given,

Equations : x + 2y + 9 = 0, 3x + 4y + 17 = 0

⇒ x + 2y + 9 = 0

⇒ x = -9 - 2y     ....(1)

Substituting value of x from equation (1) in 3x + 4y + 17 = 0, we get :

⇒ 3(-9 - 2y) + 4y + 17 = 0

⇒ -27 - 6y + 4y + 17 = 0

⇒ -10 - 2y = 0

⇒ -2y = 10

⇒ y = $-\frac{10}{2} = -5$.

Substituting value of y in equation (1), we get :

⇒ x = -9 - 2y

⇒ x = -9 - 2(-5)

⇒ x = -9 + 10

⇒ x = 1.

Hence, x = 1, y = -5.

Given,

Equations : 10x + 3y = 75, 6x - 5y = 11

⇒ 10x + 3y = 75

⇒ 10x = 75 - 3y

⇒ x = $\frac{75 - 3y}{10}$     ....(1)

Substituting value of x from equation (1) in 6x - 5y = 11, we get :

$$\begin{aligned}\Rightarrow 6\Big(\frac{75 - 3y}{10}\Big) - 5y = 11 \\[1em] \Rightarrow 3\Big(\frac{75 - 3y}{5}\Big) - 5y = 11 \\[1em] \Rightarrow \Big(\frac{225 - 9y}{5}\Big) - 5y = 11 \\[1em] \Rightarrow \Big(\frac{225 - 9y - 25y}{5}\Big) = 11 \\[1em] \Rightarrow \Big(\frac{225 - 34y}{5}\Big) = 11 \\[1em] \Rightarrow 225 - 34y = 11 \times 5\\[1em] \Rightarrow -34y = 55 - 225 \\[1em] \Rightarrow -34y = -170 \\[1em] \Rightarrow y = \frac{170}{34} \\[1em] \Rightarrow y = 5.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{75 - 3y}{10} \\[1em] \Rightarrow x = \frac{75 - 3(5)}{10} \\[1em] \Rightarrow x = \frac{75 - 15}{10} \\[1em] \Rightarrow x = \frac{60}{10} \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, x = 6, y = 5.

Given,

Equations : 7x - 2y = 20, 11x + 15y + 23 = 0

⇒ 7x - 2y = 20

⇒ 7x = 20 + 2y

⇒ x = $\frac{20 + 2y}{7}$     ....(1)

Substituting value of x from equation (1) in 11x + 15y + 23 = 0, we get :

$$\begin{aligned}\Rightarrow 11\Big(\frac{20 + 2y}{7}\Big) + 15y + 23 = 0 \\[1em] \Rightarrow \Big(\frac{220 + 22y}{7}\Big) + 15y + 23 = 0 \\[1em] \Rightarrow \Big(\frac{220 + 22y + 105y + 161}{7}\Big) = 0 \\[1em] \Rightarrow 381 + 127y = 0 \\[1em] \Rightarrow 127y = - 381 \\[1em] \Rightarrow y = \frac{-381}{127}\\[1em] \Rightarrow y = -3.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{20 + 2y}{7} \\[1em] \Rightarrow x = \frac{20 + 2(-3)}{7} \\[1em] \Rightarrow x = \frac{20 - 6}{7} \\[1em] \Rightarrow x = \frac{14}{7} \\[1em] \Rightarrow x = 2.\end{aligned}$$

Hence, x = 2, y = -3.

Given,

Equations : $\frac{7 - 4x}{3} = y$, 2x + 3y + 1 = 0

⇒ $y = \frac{7 - 4x}{3}$     ....(1)

Substituting value of y from equation (1) in 2x + 3y + 1 = 0, we get :

$\Rightarrow 2x + 3\Big(\frac{7 - 4x}{3}\Big) + 1 = 0$

⇒ 2x + (7 - 4x) + 1 = 0

⇒ 8 - 2x = 0

⇒ 2x = 8

⇒ x = $\frac{8}{2}$

⇒ x = 4.

Substituting value of x in equation (1), we get :

$$\begin{aligned}\Rightarrow y = \frac{7 - 4x}{3} \\[1em] \Rightarrow y = \frac{7 - 4(4)}{3} \\[1em] \Rightarrow y = \frac{7 - 16}{3} \\[1em] \Rightarrow y = -\frac{9}{3} \\[1em] \Rightarrow y = -3.\end{aligned}$$ Hence, x = 4, y = -3.

Given,

Equations : 4x - 3y = 8, 18x - 3y = 29

⇒ 4x - 3y = 8

⇒ 4x = 3y + 8

⇒ x = $\frac{3y + 8}{4}$     ....(1)

Substituting value of x from equation (1) in 18x - 3y = 29, we get :

$$\begin{aligned}\Rightarrow 18\Big(\frac{3y + 8}{4}\Big) - 3y = 29 \\[1em] \Rightarrow 9\Big(\frac{3y + 8}{2}\Big) - 3y = 29 \\[1em] \Rightarrow \Big(\frac{27y + 72}{2}\Big) - 3y = 29 \\[1em] \Rightarrow \Big(\frac{27y + 72 - 6y}{2}\Big) = 29 \\[1em] \Rightarrow 21y + 72 = 29 \times 2 \\[1em] \Rightarrow 21y + 72 = 58 \\[1em] \Rightarrow 21y = 58 - 72 \\[1em] \Rightarrow 21y = -14 \\[1em] \Rightarrow y = \frac{-14}{21} \\[1em] \Rightarrow y = -\frac{2}{3}.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{3y + 8}{4} \\[1em] \Rightarrow x = \frac{3 \Big(\frac{-2}{3}\Big) + 8}{4} \\[1em] \Rightarrow x = \frac{-2 + 8}{4} \\[1em] \Rightarrow x = \frac{6}{4} \\[1em] \Rightarrow x = \frac{3}{2}.\end{aligned}$$

Hence, $x = \frac{3}{2}, y = -\frac{2}{3}$.

Given,

$\frac{x + y - 8}{2} = \frac{x + 2y - 14}{3} = \frac{3x + y - 12}{11}$

Solving L.H.S. of the given equation,

$$\begin{aligned}\Rightarrow \frac{x + y - 8}{2} = \frac{x + 2y - 14}{3} \\[1em] \Rightarrow 3(x + y - 8) = 2(x + 2y - 14) \\[1em] \Rightarrow 3x + 3y - 24 = 2x + 4y - 28 \\[1em] \Rightarrow 3x + 3y - 2x - 4y = -28 + 24 \\[1em] \Rightarrow x - y = -4 \\[1em] \Rightarrow x = -4 + y \text{ ....(1)}\end{aligned}$$

Solving R.H.S. of the given equation,

$$\begin{aligned}\Rightarrow \frac{x + 2y - 14}{3} = \frac{3x + y - 12}{11} \\[1em] \Rightarrow 11(x + 2y - 14) = 3(3x + y - 12) \\[1em] \Rightarrow 11x + 22y - 154 = 9x + 3y - 36 \\[1em] \Rightarrow 11x + 22y - 9x - 3y = -36 + 154 \\[1em] \Rightarrow 2x + 19y = 118 \text{ ........(2)}\end{aligned}$$

Substituting value of x from equation (1) in (2), we get :

⇒ 2(-4 + y) + 19y = 118

⇒ -8 + 2y + 19y = 118

⇒ 21y = 118 + 8

⇒ 21y = 126

⇒ y = $\frac{126}{21} = 6$.

Substituting value of y in equation (1), we get :

⇒ x = -4 + y

⇒ x = -4 + 6

⇒ x = 2.

Hence, x = 2, y = 6.

Given,

$\frac{x}{7} + \frac{y}{3} = 5, \frac{x}{2} - \frac{y}{9} = 6$

Simplifying,

$$\begin{aligned}\Rightarrow \frac{x}{7} + \frac{y}{3} = 5 \\[1em] \Rightarrow \frac{3x + 7y}{21} = 5 \\[1em] \Rightarrow 3x + 7y = 5 \times 21 \\[1em] \Rightarrow 3x + 7y = 105 \\[1em] \Rightarrow 3x = 105 - 7y \\[1em] \Rightarrow x = \frac{105 - 7y}{3} \text{ ....(1)}\end{aligned}$$

Substituting value of x from equation (1) in $\frac{x}{2} - \frac{y}{9} = 6$, we get :

$$\begin{aligned}\Rightarrow \frac{\frac{105 - 7y}{3}}{2} - \frac{y}{9} = 6 \\[1em] \Rightarrow \frac{(105 - 7y)}{6} - \frac{y}{9} = 6 \\[1em] \Rightarrow \frac{3(105 - 7y) - 2y}{18} = 6 \\[1em] \Rightarrow \frac{315 - 21y - 2y}{18} = 6 \\[1em] \Rightarrow 315 - 23y = 6 \times 18 \\[1em] \Rightarrow 315 - 23y = 108 \\[1em] \Rightarrow 23y = 315 - 108 \\[1em] \Rightarrow 23y = 207 \\[1em] \Rightarrow y = \frac{207}{23} = 9.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{105 - 7y}{3} \\[1em] \Rightarrow x = \frac{105 - 7(9)}{3} \\[1em] \Rightarrow x = \frac{105 - 63}{3} \\[1em] \Rightarrow x = \frac{42}{3} \\[1em] \Rightarrow x = 14.\end{aligned}$$

Hence, x = 14, y = 9.

Simplifying, equation : $\frac{x}{6} + 6 = y$

$$\begin{aligned}\Rightarrow \frac{x}{6} + 6 = y \\[1em] \Rightarrow \frac{x + 36}{6} = y \\[1em] \Rightarrow x + 36 = 6y \\[1em] \Rightarrow x = 6y - 36 \text{ ....(1)}\end{aligned}$$

Substituting value of x from equation (1) in $\frac{3x}{4} = 1 + y$, we get :

$$\begin{aligned}\Rightarrow \frac{3(6y - 36)}{4} = 1 + y \\[1em] \Rightarrow \frac{18y - 108}{4} = 1 + y \\[1em] \Rightarrow 18y - 108 = 4(1 + y) \\[1em] \Rightarrow 18y - 108 = (4 + 4y) \\[1em] \Rightarrow 18y - 4y = 4 + 108 \\[1em] \Rightarrow 14y = 112 \\[1em] \Rightarrow y = \frac{112}{14} \\[1em] \Rightarrow y = 8.\end{aligned}$$

Substituting value of y in equation (1), we get :

⇒ x = 6y - 36

⇒ x = 6(8) - 36

⇒ x = 48 - 36

⇒ x = 12.

Hence, x = 12, y = 8.

Simplifying, equation : $4x + \frac{x - y}{8} = 17$

$$\begin{aligned}\Rightarrow 4x + \frac{x - y}{8} = 17 \\[1em] \Rightarrow \frac{32x + x - y}{8} = 17 \\[1em] \Rightarrow (33x - y) = 17 \times 8 \\[1em] \Rightarrow (33x - y) = 136 \\[1em] \Rightarrow 33x - y = 136 \\[1em] \Rightarrow y = 33x - 136 \text{ ....(1)}\end{aligned}$$

Substituting value of y from equation (1) in $x + 2y = \frac{y - 2}{3} - 2$, we get :

$$\begin{aligned}\Rightarrow x + 2y = \frac{y - 2}{3} - 2 \\[1em] \Rightarrow x + 2(33x - 136) = \frac{33x - 136 - 2}{3} - 2 \\[1em] \Rightarrow x + 66x - 272 = \frac{33x -138}{3} - 2 \\[1em] \Rightarrow 67x - 272 = \frac{3(11x - 46)}{3} - 2 \\[1em] \Rightarrow 67x - 272 = (11x - 46) - 2 \\[1em] \Rightarrow 67x - 272 = (11x - 48) \\[1em] \Rightarrow 67x - 11x = -48 + 272 \\[1em] \Rightarrow 56x = 224 \\[1em] \Rightarrow x = \frac{224}{56} = 4.\end{aligned}$$

Substituting value of y in equation (1), we get :

⇒ y = 33(4) - 136

⇒ y = 132 - 136

⇒ y = -4.

Hence, x = 4, y = -4.

Simplifying, equation : $\frac{x}{2} + y = \frac{4}{5}$

$$\begin{aligned}\Rightarrow \frac{x}{2} + y = \frac{4}{5} \\[1em] \Rightarrow \frac{x + 2y}{2} = \frac{4}{5} \\[1em] \Rightarrow 5(x + 2y) = 4 \times 2 \\[1em] \Rightarrow 5x + 10y = 8 \\[1em] \Rightarrow 10y = 8 - 5x \\[1em] \Rightarrow y = \frac{8 - 5x}{10} \text{ ....(1)}\end{aligned}$$

Substituting value of y from equation (1) in $x + \frac{y}{2} = \frac{7}{10}$, we get :

$$\begin{aligned}\Rightarrow x + \frac{\frac{8 - 5x}{10}}{2} = \frac{7}{10} \\[1em] \Rightarrow x + \frac{8 - 5x}{20} = \frac{7}{10} \\[1em] \Rightarrow \frac{20x + 8 - 5x}{20} = \frac{7}{10} \\[1em] \Rightarrow 15x + 8 = \frac{7}{10} \times 20 \\[1em] \Rightarrow 15x + 8 = 7 \times 2 \\[1em] \Rightarrow 15x = 14 - 8 \\[1em] \Rightarrow 15x = 6 \\[1em] \Rightarrow x = \frac{6}{15} = \frac{2}{5}.\end{aligned}$$

Substituting value of x in equation (1), we get :

$$\begin{aligned}\Rightarrow y = \frac{8 - 5x}{10} \\[1em] \Rightarrow y = \frac{8 - 5 \Big(\frac{2}{5}\Big)}{10} \\[1em] \Rightarrow y = \frac{8 - 2}{10} \\[1em] \Rightarrow y = \frac{6}{10} = \frac{3}{5}.\end{aligned}$$

Hence, $x = \frac{2}{5}, y = \frac{3}{5}$.

Simplifying equation : $\frac{7 + x}{5} - \frac{2x - y}{4} = 3y - 5$

$$\begin{aligned}\Rightarrow \frac{7 + x}{5} - \frac{2x - y}{4} = 3y - 5 \\[1em] \Rightarrow \frac{4(7 + x) - 5(2x - y)}{20} = 3y - 5 \\[1em] \Rightarrow \frac{28 + 4x - 10x + 5y}{20} = 3y - 5 \\[1em] \Rightarrow 28 + 4x - 10x + 5y = 20(3y - 5) \\[1em] \Rightarrow 28 - 6x + 5y = 60y - 100 \\[1em] \Rightarrow 60y - 5y + 6x = 28 + 100 \\[1em] \Rightarrow 6x + 55y = 128 \\[1em] \Rightarrow 6x = 128 - 55y \\[1em] \Rightarrow x = \frac{128 - 55y}{6} \text{ ....(1)}\end{aligned}$$

Simplifying equation : $\frac{4x - 3}{6} + \frac{5y - 7}{2} = 18 - 5x$

$$\begin{aligned}\Rightarrow \frac{4x - 3}{6} + \frac{5y - 7}{2} = 18 - 5x \\[1em] \Rightarrow \frac{2(4x - 3) + 6(5y - 7)}{12} = 18 - 5x \\[1em] \Rightarrow \frac{8x - 6 + 30y - 42}{12} = 18 - 5x \\[1em] \Rightarrow 8x - 6 + 30y - 42 = 12(18 - 5x) \\[1em] \Rightarrow 8x + 30y - 48 = 216 - 60x \\[1em] \Rightarrow 8x + 30y + 60x = 216 + 48 \\[1em] \Rightarrow 68x + 30y = 264 \text{ ....(2) }\end{aligned}$$

Substituting value of x from equation (1) in 68x + 30y = 264, we get :

$$\begin{aligned}\Rightarrow 68\Big(\frac{128 - 55y}{6}\Big) + 30y = 264 \\[1em] \Rightarrow 34\Big(\frac{128 - 55y}{3}\Big) + 30y = 264 \\[1em] \Rightarrow \Big(\frac{4352 - 1870y}{3}\Big) + 30y = 264 \\[1em] \Rightarrow \Big(\frac{4352 - 1870y + 90y}{3}\Big) = 264 \\[1em] \Rightarrow (4352 - 1780y) = 264 \times 3 \\[1em] \Rightarrow (4352 - 1780y) = 792 \\[1em] \Rightarrow 1780y = 4352 - 792 \\[1em] \Rightarrow 1780y = 3560 \\[1em] \Rightarrow y = \frac{3560}{1780} = 2.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{128 - 55y}{6} \\[1em] \Rightarrow x = \frac{128 - 55(2)}{6} \\[1em] \Rightarrow x = \frac{128 - 110}{6} \\[1em] \Rightarrow x = \frac{18}{6} \\[1em] \Rightarrow x = 3.\end{aligned}$$

Hence, x = 3, y = 2.

Given,

Equations:

$$\begin{aligned}\Rightarrow 4x + \frac{6}{y} = 15 \text{ ....(1)} \\[1em] \Rightarrow 3x - \frac{4}{y} = 7 \text{ ....(2)}\end{aligned}$$

Multiplying equation (1) by 4, we get :

$$\begin{aligned}\Rightarrow 4\Big(4x + \frac{6}{y}\Big) = 15 \times 4 \\[1em] \Rightarrow 4\Big(4x + \frac{6}{y}\Big) = 15 \times 4 \\[1em] \Rightarrow 16x + \frac{24}{y} = 60\text{ ....(3)}\end{aligned}$$

Multiplying equation (2) by 6, we get :

$$\begin{aligned}\Rightarrow 6\Big(3x - \frac{4}{y}\Big) = 7 \times 6 \\[1em] \Rightarrow 18x - \frac{24}{y} = 42\text{ ....(4)}\end{aligned}$$

Adding equation (3) and (4), we get:

$$\begin{aligned}\Rightarrow \Big(16x + \frac{24}{y}\Big) + \Big(18x - \frac{24}{y}\Big) = 60 + 42 \\[1em] \Rightarrow 18x + \frac{24}{y} + 16x - \frac{24}{y} = 102 \\[1em] \Rightarrow 34x = 102 \\[1em] \Rightarrow x = \frac{102}{34} \\[1em] \Rightarrow x = 3.\end{aligned}$$

Substituting value of x in equation (1), we get :

$$\begin{aligned}\Rightarrow 4(3) + \frac{6}{y} = 15 \\[1em] \Rightarrow 12 + \frac{6}{y} = 15 \\[1em] \Rightarrow \frac{6}{y} = 15 - 12 \\[1em] \Rightarrow \frac{6}{y} = 3 \\[1em] \Rightarrow y = \frac{6}{3} \\[1em] \Rightarrow y = 2.\end{aligned}$$

Hence, x = 3, y = 2.

Equations:

$$\begin{aligned}\Rightarrow 5x - 9 = \frac{1}{y} \text{ ....(1)} \\[1em] \Rightarrow x + \frac{1}{y} = 3 \\[1em] \Rightarrow \frac{1}{y} = 3 - x \text{ ....(2)}\end{aligned}$$

From equation (1) and (2), we get :

$$\begin{aligned}\Rightarrow 5x - 9 = 3 - x \\[1em] \Rightarrow 5x + x = 3 + 9 \\[1em] \Rightarrow 6x = 12 \\[1em] \Rightarrow x = \frac{12}{6} \\[1em] \Rightarrow x = 2.\end{aligned}$$

Substituting value of x in equation (2), we get :

$$\begin{aligned}\Rightarrow \frac{1}{y} = 3 - 2 \\[1em] \Rightarrow \frac{1}{y} = 1 \\[1em] \Rightarrow y = \frac{1}{1} \\[1em] \Rightarrow y = 1.\end{aligned}$$

Hence, x = 2, y = 1.

Given,

Equations:

$\frac{2}{x} + \frac{2}{3y} = \frac{1}{6} \text{ ....(1) }$,

$\frac{3}{x} + \frac{2}{y} = 0 \text{ ....(2) }$

Multiplying equation (1) by 3, we get:

$$\begin{aligned}\Rightarrow 3\Big(\frac{2}{x} + \frac{2}{3y}\Big) = \frac{1}{6} \times 3 \\[1em] \Rightarrow \Big(\frac{6}{x} + \frac{2}{y}\Big) = \frac{1}{2} \text{ ....(3) }\end{aligned}$$

Multiplying equation (2) by 2, we get:

$$\begin{aligned}\Rightarrow 2\Big(\frac{3}{x} + \frac{2}{y} \Big) = 0 \times 2 \\[1em] \Rightarrow \Big(\frac{6}{x} + \frac{4}{y}\Big) = 0 \text{ ....(4) }\end{aligned}$$

Subtracting equation (3) from (4), we get:

$$\begin{aligned}\Rightarrow \Big(\frac{6}{x} + \frac{4}{y}\Big) - \Big(\frac{6}{x} + \frac{2}{y}\Big) = 0 - \frac{1}{2} \\[1em] \Rightarrow \Big(\frac{6}{x} + \frac{4}{y} - \frac{6}{x} - \frac{2}{y}\Big) = -\frac{1}{2} \\[1em] \Rightarrow \Big(\frac{4}{y} - \frac{2}{y}\Big) = -\frac{1}{2} \\[1em] \Rightarrow \Big(\frac{2}{y}\Big) = -\frac{1}{2} \\[1em] \Rightarrow y = \frac{2 \times 2}{-1} \\[1em] \Rightarrow y = -4.\end{aligned}$$

Substituting value of y in equation (1),

$$\begin{aligned}\Rightarrow \frac{2}{x} + \frac{2}{3(-4)} = \frac{1}{6} \\[1em] \Rightarrow \frac{2}{x} - \frac{1}{6} = \frac{1}{6} \\[1em] \Rightarrow \frac{2}{x} = \frac{1}{6} + \frac{1}{6} \\[1em] \Rightarrow \frac{2}{x} = \frac{2}{6} \\[1em] \Rightarrow \frac{2}{x} = \frac{1}{3} \\[1em] \Rightarrow x = 2 \times 3 \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, x = 6, y = -4.

Given,

Equations:

$\frac{3}{2x} + \frac{2}{3y} = 5 \text{ ....(1) },$

$\frac{5}{x} - \frac{3}{y} = 1 \text{ ....(2) }$

Multiplying equation (1) by 9, we get:

$$\begin{aligned}\Rightarrow 9\Big(\frac{3}{2x} + \frac{2}{3y}\Big) = 5 \times 9 \\[1em] \Rightarrow \Big(\frac{27}{2x} + \frac{6}{y}\Big) = 45 \text{ ..........(3) }\end{aligned}$$

Multiplying equation (2) by 2, we get:

$$\begin{aligned}\Rightarrow 2\Big(\frac{5}{x} - \frac{3}{y}\Big) = 1 \times 2 \\[1em] \Rightarrow \Big(\frac{10}{x} - \frac{6}{y}\Big) = 2 \text{ .........(4) }\end{aligned}$$

Adding equation (3) from (4), we get:

$$\begin{aligned}\Rightarrow \Big(\frac{27}{2x} + \frac{6}{y}\Big) + \Big(\frac{10}{x} - \frac{6}{y}\Big) = 45 + 2 \\[1em] \Rightarrow \Big(\frac{27}{2x} + \frac{10}{x}\Big) = 47 \\[1em] \Rightarrow \Big(\frac{27 + 20}{2x}\Big) = 47 \\[1em] \Rightarrow \Big(\frac{47}{2x}\Big) = 47 \\[1em] \Rightarrow 2x = \frac{47}{47} \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow x = \frac{1}{2}.\end{aligned}$$

Substituting value of x in equation (1),

$$\begin{aligned}\Rightarrow \frac{3}{2\Big(\frac{1}{2}\Big)} + \frac{2}{3y} = 5 \\[1em] \Rightarrow 3 + \frac{2}{3y} = 5 \\[1em] \Rightarrow \frac{2}{3y} = 5 - 3 \\[1em] \Rightarrow \frac{2}{3y} = 2 \\[1em] \Rightarrow y = \frac{2}{3 \times 2} \\[1em] \Rightarrow y = \frac{1}{3}.\end{aligned}$$

Hence, $x = \frac{1}{2}, y = \frac{1}{3}$.

Given,

Equations :

⇒ x + y = 2xy     ....(1)

⇒ x - y = 6xy     ....(2)

Adding equations (1) and (2), we get :

⇒ x + y + (x - y) = 2xy + 6xy

⇒ x + y + x - y = 8xy

⇒ 2x = 8xy

⇒ 2 = 8y

⇒ y = $\frac{2}{8} = \frac{1}{4}$.

Substituting $y = \frac{1}{4}$ in equation (1), we get :

$$\begin{aligned}\Rightarrow x + y = 2xy \\[1em] \Rightarrow x + \frac{1}{4} = 2 \times x \times \Big(\frac{1}{4}\Big) \\[1em] \Rightarrow x + \frac{1}{4} = \frac{x}{2} \\[1em] \Rightarrow x - \frac{x}{2} = \frac{-1}{4} \\[1em] \Rightarrow \frac{x}{2} = \frac{-1}{4} \\[1em] \Rightarrow x = -\frac{2}{4} \\[1em] \Rightarrow x = -\frac{1}{2}.\end{aligned}$$

Hence, $x = -\frac{1}{2}, y = \frac{1}{4}$.

Given,

Equations:

$\frac{3}{x + y} + \frac{2}{x - y} = 3$ ........(1)

$\frac{2}{x + y} + \frac{3}{x - y} = \frac{11}{3}$ ..........(2)

Multiplying equation (1) by 2, we get:

$$\begin{aligned}\Rightarrow 2\Big(\frac{3}{x + y} + \frac{2}{x - y}\Big) = 3 \times 2 \\[1em] \Rightarrow \Big(\frac{6}{x + y} + \frac{4}{x - y}\Big) = 6 \text{ .........(3) }\end{aligned}$$

Multiplying equation (2) by 3, we get:

$$\begin{aligned}\Rightarrow 3\Big(\frac{2}{x + y} + \frac{3}{x - y}\Big) = \frac{11}{3} \times 3 \\[1em] \Rightarrow \Big(\frac{6}{x + y} + \frac{9}{x - y}\Big) = 11 \text{ .........(4) }\end{aligned}$$

Subtracting equation (3) from (4), we get:

$$\begin{aligned}\Rightarrow \Big(\frac{6}{x + y} + \frac{9}{x - y}\Big) - \Big(\frac{6}{x + y} + \frac{4}{x - y}\Big) = 11 - 6 \\[1em] \Rightarrow \Big(\frac{6}{x + y} + \frac{9}{x - y} - \frac{6}{x + y} - \frac{4}{x - y}\Big) = 5 \\[1em] \Rightarrow \Big(\frac{9}{x - y} - \frac{4}{x - y}\Big) = 5 \\[1em] \Rightarrow \Big(\frac{5}{x - y}\Big) = 5 \\[1em] \Rightarrow 5 = 5(x - y) \\[1em] \Rightarrow x - y = 1 \text{ .........(5) }\end{aligned}$$

Substituting value of x - y from equation (5) in equation (1), we get:

$$\begin{aligned}\Rightarrow \frac{3}{x + y} + \frac{2}{1} = 3 \\[1em] \Rightarrow \frac{3}{x + y} + 2 = 3 \\[1em] \Rightarrow \frac{3}{x + y} = 3 - 2 \\[1em] \Rightarrow \frac{3}{x + y} = 1 \\[1em] \Rightarrow x + y = 3 \text{ .........(6) }\end{aligned}$$

Adding equations (5) and (6), we get:

$$\begin{aligned}\Rightarrow x + y + x - y = 3 + 1 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \frac{4}{2} = 2.\end{aligned}$$

Substituting value of x in equation (6), we get:

⇒ x + y = 3

⇒ 2 + y = 3

⇒ y = 3 - 2

⇒ y = 1.

Hence, x = 2, y = 1.

Given,

Equations:

$\frac{22}{x + y} + \frac{15}{x - y} = 5$ ..........(1)

$\frac{55}{x + y} + \frac{40}{x - y} = 13$ ..........(2)

Multiplying equation (1) by 5, we get:

$$\begin{aligned}\Rightarrow 5\Big(\frac{22}{x + y} + \frac{15}{x - y}\Big) = 5 \times 5 \\[1em] \Rightarrow \Big(\frac{110}{x + y} + \frac{75}{x - y}\Big) = 25 \text{ ..........(3) }\end{aligned}$$

Multiplying equation (2) by 2, we get:

$$\begin{aligned}\Rightarrow 2\Big(\frac{55}{x + y} + \frac{40}{x - y}\Big) = 13 \times 2 \\[1em] \Rightarrow \Big(\frac{110}{x + y} + \frac{80}{x - y}\Big) = 26 \text{ ..........(4) }\end{aligned}$$

Subtracting equation (3) from (4), we get:

$$\begin{aligned}\Rightarrow \Big(\frac{110}{x + y} + \frac{80}{x - y}\Big) - \Big(\frac{110}{x + y} + \frac{75}{x - y}\Big) = 26 - 25 \\[1em] \Rightarrow \Big(\frac{110}{x + y} + \frac{80}{x - y} - \frac{110}{x + y} - \frac{75}{x - y}\Big) = 1 \\[1em] \Rightarrow \Big(\frac{80 - 75}{x - y}\Big) = 1 \\[1em] \Rightarrow \Big(\frac{5}{x - y}\Big) = 1 \\[1em] \Rightarrow 5 = (x - y) \\[1em] \Rightarrow x - y = 5 \text{ .........(5) }\end{aligned}$$

Substituting value of x - y from equation (5) in equation (1), we get:

$$\begin{aligned}\Rightarrow \frac{22}{x + y} + \frac{15}{x - y} = 5 \\[1em] \Rightarrow \frac{22}{x + y} + \frac{15}{5} = 5 \\[1em] \Rightarrow \frac{22}{x + y} + 3 = 5 \\[1em] \Rightarrow \frac{22}{x + y} = 5 - 3 \\[1em] \Rightarrow \frac{22}{x + y} = 2 \\[1em] \Rightarrow x + y = \frac{22}{2} \\[1em] \Rightarrow x + y = 11 \text{ .......(6) }\end{aligned}$$

Adding equations (5) and (6), we get:

$$\begin{aligned}\Rightarrow x + y + x - y = 11 + 5 \\[1em] \Rightarrow 2x = 16 \\[1em] \Rightarrow x = \frac{16}{2} = 8.\end{aligned}$$

Substituting value of x in equation (6), we get:

⇒ x + y = 11

⇒ 8 + y = 11

⇒ y = 11 - 8

⇒ y = 3

Hence, x = 8, y = 3.

Given,

Equations :

103x + 51y = 617     .......(1),

97x + 49y = 583     .......(2)

Adding equations (1) and (2),

⇒ 103x + 51y + (97x + 49y) = 617 + 583

⇒ 103x + 51y + 97x + 49y = 617 + 583

⇒ 200x + 100y = 1200

⇒ 100(2x + y) = 1200

⇒ 2x + y = $\frac{1200}{100}$

⇒ 2x + y = 12     ........(3)

Subtracting equation (2) from (1),

⇒ 103x + 51y - (97x + 49y) = 617 - 583

⇒ 103x + 51y - 97x - 49y = 34

⇒ 6x + 2y = 34

⇒ 2(3x + y) = 34

⇒ 3x + y = $\frac{34}{2}$

⇒ 3x + y = 17     ........(4)

Subtracting equation 4 from 3,

⇒ 2x + y - (3x + y) = 12 - 17

⇒ 2x + y - 3x - y = -5

⇒ -x = -5

⇒ x = 5.

Substituting value of x in equation 1,

⇒ 103x + 51y = 617

⇒ 103(5) + 51y = 617

⇒ 515 + 51y = 617

⇒ 51y = 617 - 515

⇒ 51y = 102

⇒ y = $\frac{102}{51}$ = 2.

Hence, x = 5, y = 2.

Given,

Equations :

23x − 29y = 98     .........(1),

29x − 23y = 110     ........(2)

Adding equations 1 and 2,

⇒ 23x − 29y + (29x − 23y) = 98 + 110

⇒ 23x − 29y + 29x − 23y = 98 + 110

⇒ 52x − 52y = 208

⇒ 52(x - y) = 208

⇒ (x - y) = $\frac{208}{52}$

⇒ x - y = 4     .........(3)

Subtracting equation 2 from 1,

⇒ 23x − 29y - (29x − 23y) = 98 - 110

⇒ 23x − 29y - 29x + 23y = 98 - 110

⇒ -6x - 6y = -12

⇒ -6(x + y) = -12

⇒ x + y = $\frac{-12}{-6}$

⇒ x + y = 2     .......(4)

Adding equation 4 and 5,

⇒ x + y + x - y = 2 + 4

⇒ 2x = 6

⇒ $x = \frac{6}{2}$

⇒ x = 3.

Substituting value of x in equation 1,

⇒ 23x − 29y = 98

⇒ 23(3) − 29y = 98

⇒ 69 − 29y = 98

⇒ −29y = 98 − 69

⇒ −29y = 29

⇒ y = $\frac{29}{-29}$ = -1.

Hence, x = 3, y = -1.

Substituting $\frac{1}{x} = m \text{ and } \frac{1}{y} = n$ in $\frac{a}{x} - \frac{b}{y} = 0$,

⇒ am - bn = 0

⇒ am = bn

⇒ m = $\frac{bn}{a}$     ........(1)

Substituting $\frac{1}{x} = m \text{ and } \frac{1}{y} = n$ in $\frac{ab^2}{x} + \frac{a^2b}{y} = (a^2 + b^2)$

⇒ ab²m + a²bn = a² + b²     .........(2)

Substituting value of m from equation (1) in (2), we get :

⇒ ab² $\Big(\frac{bn}{a}\Big)$ + a²bn = a² + b²

⇒ b³n + a²bn = a² + b²

⇒ bn(b² + a²) = a² + b²

⇒ bn = $\frac{(a^2 + b^2)}{(a^2 + b^2)}$

⇒ bn = 1

⇒ n = $\frac{1}{b}$

Substituting value of n in equation (1), we get:

⇒ m = $\frac{b}{a} \times \frac{1}{b}$

⇒ m = $\frac{1}{a}$

$$\begin{aligned}\Rightarrow \frac{1}{x} = m \\[1em] \Rightarrow \frac{1}{x} = \frac{1}{a} \\[1em] \Rightarrow x = a \\[1em] \Rightarrow \frac{1}{y} = n \\[1em] \Rightarrow \frac{1}{y} = \frac{1}{b} \\[1em] \Rightarrow y = b.\end{aligned}$$

Hence, x = a, y = b.

Given,

Equations :

2x + y = 32     .........(1),

3x + 4y = 68     .........(2)

Multiplying equation (1) by 4, we get :

⇒ 4(2x + y) = 32 × 4

⇒ 8x + 4y = 128     .........(3)

Subtracting equation (2) from (3), we get :

⇒ 8x + 4y - (3x + 4y) = 128 - 68

⇒ 8x + 4y - 3x - 4y = 60

⇒ 5x = 60

⇒ x = $\frac{60}{5}$

⇒ x = 12.

Substituting value of x in equation (2), we get :

⇒ 3x + 4y = 68

⇒ 3 × 12 + 4y = 68

⇒ 36 + 4y = 68

⇒ 4y = 68 - 36

⇒ 4y = 32

⇒ y = $\frac{32}{4}$

⇒ y = 8.

Substituting value of x and y in $\frac{x}{y}$, we get :

$\Rightarrow \frac{x}{y} = \frac{12}{8} = \frac{3}{2}$.

Hence, $\frac{x}{y} = \frac{3}{2}$.

Given,

In an equilateral triangle, all three sides are equal.

∴ x + 3y = 3x + 2y - 2 = 4x + $\frac{y}{2}$ + 1

Solving L.H.S of the above equation, we get :

⇒ x + 3y = 3x + 2y - 2

⇒ x - 3x + 3y - 2y = -2

⇒ -2x + y = -2

⇒ y = 2x - 2     .....(1)

Solving R.H.S of the above equation, we get :

$$\begin{aligned}\Rightarrow 3x + 2y - 2 = 4x + \frac{y}{2} + 1 \\[1em] \Rightarrow 3x + 2y = 4x + \frac{y}{2} + 1 + 2 \\[1em] \Rightarrow 3x - 4x + 2y - \frac{y}{2} = 3 \\[1em] \Rightarrow -x + \frac{4y - y}{2} = 3 \\[1em] \Rightarrow -x + \frac{3y}{2} = 3 \\[1em] \Rightarrow \frac{3y - 2x}{2} = 3 \\[1em] \Rightarrow 3y - 2x = 3 \times 2 \\[1em] \Rightarrow 3y - 2x = 6 \text{ .........(2)}\end{aligned}$$

Substituting value of y from equation (1) in (2), we get :

⇒ 3(2x - 2) - 2x = 6

⇒ 6x - 6 - 2x = 6

⇒ 4x - 6 = 6

⇒ 4x = 6 + 6

⇒ 4x = 12

⇒ x = $\frac{12}{4}$

⇒ x = 3.

Substituting value of x in equation (1), we get :

⇒ y = 2(3) - 2

⇒ y = 6 - 2

⇒ y = 4.

Substituting value of x and y in x + 3y, we get :

⇒ x + 3y = 3 + 3(4) = 3 + 12 = 15 cm.

Since, all sides are equal.

Hence, the length of each side is 15 cm.

Given,

Equations:

⇒ 2x − 5y + 8 = 0

⇒ x − 4y + 7 = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(-5) \times (7) - (-4) \times (8)} = \frac{y}{(8) \times (1) - (7) \times (2)} = \frac{1}{(2) \times (-4) - (1) \times (-5)} \\[1em] \Rightarrow \frac{x}{(-35) + 32} = \frac{y}{8 - 14} = \frac{1}{-8 + 5} \\[1em] \Rightarrow \frac{x}{-3} = \frac{y}{-6} = \frac{1}{-3} \\[1em] \Rightarrow \frac{x}{-3} = \frac{1}{-3} \text{ and } \frac{y}{-6} = \frac{1}{-3} \\[1em] \Rightarrow x = \frac{-3}{-3} \text{ and } y = \frac{-6}{-3} \\[1em] \Rightarrow x = 1 \text{ and } y = 2.\end{aligned}$$

Hence, x = 1 and y = 2.

Given,

Equations :

⇒ 5x − 4y + 2 = 0

⇒ 2x + 3y - 13 = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(-4) \times (-13) - (3) \times (2)} = \frac{y}{(2) \times (2) - (-13) \times (5)} = \frac{1}{(5) \times (3) - (2) \times (-4)} \\[1em] \Rightarrow \frac{x}{(52) - 6} = \frac{y}{4 + 65} = \frac{1}{15 + 8} \\[1em] \Rightarrow \frac{x}{46} = \frac{y}{69} = \frac{1}{23} \\[1em] \Rightarrow \frac{x}{46} = \frac{1}{23} \text{ and } \frac{y}{69} = \frac{1}{23} \\[1em] \Rightarrow x = \frac{46}{23} \text{ and } y = \frac{69}{23} \\[1em] \Rightarrow x = 2 \text{ and } y = 3.\end{aligned}$$

Hence, x = 2 and y = 3.

Given,

Equations:

⇒ 3x − 5y - 19 = 0

⇒ 7x − 3y - 1 = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(-5) \times (-1) - (-3) \times (-19)} = \frac{y}{(-19) \times (7) - (-1) \times (3)} = \frac{1}{(3) \times (-3) - (7) \times (-5)} \\[1em] \Rightarrow \frac{x}{5 - 57} = \frac{y}{-133 + 3} = \frac{1}{-9 + 35} \\[1em] \Rightarrow \frac{x}{-52} = \frac{y}{-130} = \frac{1}{26} \\[1em] \Rightarrow \frac{x}{-52} = \frac{1}{26} \text{ and } \frac{y}{-130} = \frac{1}{26} \\[1em] \Rightarrow x = \frac{-52}{26} \text{ and } y = \frac{-130}{26} \\[1em] \Rightarrow x = -2 \text{ and } y = -5.\end{aligned}$$

Hence, x = -2 and y = -5.

Given,

Equations:

⇒ 2x + 3y - 17 = 0

⇒ 3x - 2y - 6 = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(3) \times (-6) - (-2) \times (-17)} = \frac{y}{(-17) \times (3) - (-6) \times (2)} = \frac{1}{(2) \times (-2) - (3) \times (3)} \\[1em] \Rightarrow \frac{x}{(-18) - 34} = \frac{y}{-51 + 12} = \frac{1}{-4 - 9} \\[1em] \Rightarrow \frac{x}{-52} = \frac{y}{-39} = \frac{1}{-13} \\[1em] \Rightarrow \frac{x}{-52} = \frac{1}{-13} \text{ and } \frac{y}{-39} = \frac{1}{-13} \\[1em] \Rightarrow x = \frac{-52}{-13} \text{ and } y = \frac{-39}{-13} \\[1em] \Rightarrow x = 4 \text{ and } y = 3.\end{aligned}$$

Hence, x = 4 and y = 3.

Given,

Equations:

⇒ x + 2y + 1 = 0

⇒ 2x − 3y - 12 = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(2) \times (-12) - (-3) \times (1)} = \frac{y}{(1) \times (2) - (-12) \times (1)} = \frac{1}{(1) \times (-3) - (2) \times (2)} \\[1em] \Rightarrow \frac{x}{(-24) + 3} = \frac{y}{2 + 12} = \frac{1}{-3 - 4} \\[1em] \Rightarrow \frac{x}{-21} = \frac{y}{14} = \frac{1}{-7} \\[1em] \Rightarrow \frac{x}{-21} = \frac{1}{-7} \text{ and } \frac{y}{14} = \frac{1}{-7} \\[1em] \Rightarrow x = \frac{-21}{-7} \text{ and } y = \frac{14}{-7} \\[1em] \Rightarrow x = 3 \text{ and } y = -2.\end{aligned}$$

Hence, x = 3 and y = -2.

Given,

Equations:

⇒ 2x + 5y - 1 = 0

⇒ 2x + 3y - 3 = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(5) \times (-3) - (3) \times (-1)} = \frac{y}{(-1) \times (2) - (-3) \times (2)} = \frac{1}{(2) \times (3) - (2) \times (5)} \\[1em] \Rightarrow \frac{x}{(-15) + 3} = \frac{y}{-2 + 6} = \frac{1}{6 - 10} \\[1em] \Rightarrow \frac{x}{-12} = \frac{y}{4} = \frac{1}{-4} \\[1em] \Rightarrow \frac{x}{-12} = \frac{1}{-4} \text{ and } \frac{y}{4} = \frac{1}{-4} \\[1em] \Rightarrow x = \frac{-12}{-4} \text{ and } y = \frac{4}{-4} \\[1em] \Rightarrow x = 3 \text{ and } y = -1.\end{aligned}$$

Hence, x = 3 and y = -1.

Given,

Equations:

⇒ 8x − 3y - 12 = 0

⇒ 5x - 2y - 7 = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(-3) \times (-7) - (-2) \times (-12)} = \frac{y}{(-12) \times (5) - (-7) \times (8)} = \frac{1}{(8) \times (-2) - (5) \times (-3)} \\[1em] \Rightarrow \frac{x}{21 - 24} = \frac{y}{-60 + 56} = \frac{1}{-16 + 15} \\[1em] \Rightarrow \frac{x}{-3} = \frac{y}{-4} = \frac{1}{-1} \\[1em] \Rightarrow \frac{x}{-3} = \frac{1}{-1} \text{ and } \frac{y}{-4} = \frac{1}{-1} \\[1em] \Rightarrow x = \frac{-3}{-1} \text{ and } y = \frac{-4}{-1} \\[1em] \Rightarrow x = 3 \text{ and } y = 4.\end{aligned}$$

Hence, x = 3 and y = 4.

Given,

Equations:

⇒ 7x − 2y - 20 = 0

⇒ 11x + 15y + 23 = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(-2) \times (23) - (15) \times (-20)} = \frac{y}{(-20) \times (11) - (23) \times (7)} = \frac{1}{(7) \times (15) - (11) \times (-2)} \\[1em] \Rightarrow \frac{x}{(-46) + 300} = \frac{y}{-220 - 161} = \frac{1}{105 + 22} \\[1em] \Rightarrow \frac{x}{254} = \frac{y}{-381} = \frac{1}{127} \\[1em] \Rightarrow \frac{x}{254} = \frac{1}{127} \text{ and } \frac{y}{-381} = \frac{1}{127} \\[1em] \Rightarrow x = \frac{254}{127} \text{ and } y = \frac{-381}{127} \\[1em] \Rightarrow x = 2 \text{ and } y = -3.\end{aligned}$$

Hence, x = 2 and y = -3.

Given,

Equations:

⇒ ax + by - (a − b) = 0

⇒ bx − ay - (a + b) = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(b) \times -(a + b) - [(-a) \times -(a - b)]} = \frac{y}{-(a - b) \times (b) - [-(a + b)] \times (a)} = \frac{1}{(a) \times (-a) - (b) \times (b)}\\[1em] \Rightarrow \frac{x}{(b) \times (-a - b) - [(-a) \times (-a + b)]} = \frac{y}{-(a - b) \times (b) - (-a - b) \times (a)} = \frac{1}{(a) \times (-a) - (b) \times (b)}\\[1em] \Rightarrow \frac{x}{-ab - b^2 - (a^2 - ab)} = \frac{y}{-ab + b^2 - (-a^2 - ab)} = \frac{1}{-a^2 - b^2} \\[1em] \Rightarrow \frac{x}{(-ab - b^2 - a^2 + ab)} = \frac{y}{-ab + b^2 + a^2 + ab} = \frac{1}{-a^2 - b^2} \\[1em] \Rightarrow \frac{x}{-a^2 - b^2} = \frac{y}{a^2 + b^2} = \frac{1}{-a^2 - b^2} \\[1em] \Rightarrow \frac{x}{-a^2 - b^2} = \frac{1}{-a^2 - b^2} \text{ and } \frac{y}{a^2 + b^2} = \frac{1}{-a^2 - b^2} \\[1em] \Rightarrow x = \frac{-a^2 - b^2}{-a^2 - b^2} \text{ and } y = \frac{a^2 + b^2}{-a^2 - b^2} = \frac{a^2+ b^2}{-(a^2 + b^2)} \\[1em] \Rightarrow x = 1 \text{ and } y = -1.\end{aligned}$$

Hence, x = 1 and y = -1.

Given,

Equations :

⇒ 3x + 2y + 25 = 0

⇒ 2x + y + 10 = 0

By cross-multiplication method,

$$\begin{aligned}\Rightarrow \frac{x}{(2) \times (10) - (1) \times (25)} = \frac{y}{(25) \times (2) - (10) \times (3)} = \frac{1}{(3) \times (1) - (2) \times (2)} \\[1em] \Rightarrow \frac{x}{20 - 25} = \frac{y}{50 - 30} = \frac{1}{3 - 4} \\[1em] \Rightarrow \frac{x}{-5} = \frac{y}{20} = \frac{1}{-1} \\[1em] \Rightarrow \frac{x}{-5} = \frac{1}{-1} \text{ and } \frac{y}{20} = \frac{1}{-1} \\[1em] \Rightarrow x = \frac{-5}{-1} \text{ and } y = \frac{20}{-1} \\[1em] \Rightarrow x = 5 \text{ and } y = -20.\end{aligned}$$

Hence, x = 5 and y = -20.

Substituting $\frac{1}{x}= a, \frac{1}{y} = b$ in $\frac{5}{x} - \frac{4}{y} + 2 = 0$, we get:

⇒ 5a - 4b + 2 = 0     ..........(1)

Substituting $\frac{1}{x} = a, \frac{1}{y} = b$ in $\frac{2}{x} + \frac{3}{y} = 13$, we get :

⇒ 2a + 3b = 13

⇒ 2a + 3b - 13 = 0     ........(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$$\begin{aligned}\Rightarrow \frac{a}{(-4) \times (-13) - (3) \times (2)} = \frac{b}{(2) \times (2) - (-13) \times (5)} = \frac{1}{(5) \times (3) - (2) \times (-4)} \\[1em] \Rightarrow \frac{a}{52 - 6} = \frac{b}{4 + 65} = \frac{1}{15 + 8} \\[1em] \Rightarrow \frac{a}{46} = \frac{b}{69} = \frac{1}{23} \\[1em] \Rightarrow \frac{a}{46} = \frac{1}{23} \text{ and } \frac{b}{69} = \frac{1}{23} \\[1em] \Rightarrow a = \frac{46}{23} \text{ and } b = \frac{69}{23} \\[1em] \Rightarrow a = 2 \text{ and } b = 3.\end{aligned}$$

Now we have a = 2 and b = 3,

$$\begin{aligned}\Rightarrow \frac{1}{x} = a \\[1em] \Rightarrow \frac{1}{x} = 2 \\[1em] \Rightarrow x = \frac{1}{2} \\[1em] \Rightarrow \frac{1}{y} = b \\[1em] \Rightarrow \frac{1}{y} = 3 \\[1em] \Rightarrow y = \frac{1}{3}.\end{aligned}$$

Hence, $x = \frac{1}{2}$ and $y = \frac{1}{3}$.

Substituting $\frac{1}{x}= a, \frac{1}{y} = b$ in $\frac{1}{x} + \frac{1}{y} = 7$, we get :

⇒ a + b = 7

⇒ a + b - 7 = 0     .........(1)

Substituting $\frac{1}{x} = a, \frac{1}{y} = b$ in $\frac{2}{x} + \frac{3}{y} = 17$, we get :

⇒ 2a + 3b = 17

⇒ 2a + 3b - 17 = 0     .........(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$$\begin{aligned}\Rightarrow \frac{a}{(1) \times (-17) - (3) \times (-7)} = \frac{b}{(-7) \times (2) - (-17) \times (1)} = \frac{1}{(1) \times (3) - (2) \times (1)} \\[1em] \Rightarrow \frac{a}{(-17) + 21} = \frac{b}{-14 + 17} = \frac{1}{3 - 2} \\[1em] \Rightarrow \frac{a}{4} = \frac{b}{3} = \frac{1}{1} \\[1em] \Rightarrow \frac{a}{4} = 1 \text{ and } \frac{b}{3} = \frac{1}{1} \\[1em] \Rightarrow a = \frac{4}{1} \text{ and } b = \frac{3}{1} \\[1em] \Rightarrow a = 4 \text{ and } b = 3.\end{aligned}$$

Now we have a = 4 and b = 3,

$$\begin{aligned}\Rightarrow \frac{1}{x} = a \\[1em] \Rightarrow \frac{1}{x} = 4 \\[1em] \Rightarrow x = \frac{1}{4} \\[1em] \Rightarrow \frac{1}{y} = b \\[1em] \Rightarrow \frac{1}{y} = 3 \\[1em] \Rightarrow y = \frac{1}{3}.\end{aligned}$$

Hence, $x = \frac{1}{4}$ and $y = \frac{1}{3}$.

Substituting $\frac{1}{x + y}= a, \frac{1}{x - y} = b$ in $\frac{10}{x + y} + \frac{2}{x - y} = 4$, we get:

⇒ 10a + 2b = 4

⇒ 10a + 2b - 4 = 0     .....(1)

Substituting $\frac{1}{x + y}= a, \frac{1}{x - y} = b$ in $\frac{15}{x + y} - \frac{5}{x - y} + 2 = 0$, we get:

⇒ 15a - 5b + 2 = 0     ....(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$$\begin{aligned}\Rightarrow \frac{a}{(2) \times (2) - (-5) \times (-4)} = \frac{b}{(-4) \times (15) - (2) \times (10)} = \frac{1}{(10) \times (-5) - (15) \times (2)} \\[1em] \Rightarrow \frac{a}{(4) - 20} = \frac{b}{-60 - 20} = \frac{1}{-50 - 30} \\[1em] \Rightarrow \frac{a}{-16} = \frac{b}{-80} = \frac{1}{-80} \\[1em] \Rightarrow \frac{a}{-16} = \frac{1}{-80} \text{ and } \frac{b}{-80} = \frac{1}{-80} \\[1em] \Rightarrow a = \frac{-16}{-80} \text{ and } b = \frac{-80}{-80} \\[1em] \Rightarrow a = \frac{1}{5} \text{ and } b = 1.\end{aligned}$$

Now we have a = $\frac{1}{5}$ and b = 1,

$$\begin{aligned}\Rightarrow \frac{1}{x + y} = a \\[1em] \Rightarrow \frac{1}{x + y} = \frac{1}{5} \\[1em] \Rightarrow x + y = 5 \text{ ..........(3)} \\[1em] \Rightarrow \frac{1}{x - y} = b \\[1em] \Rightarrow \frac{1}{x - y} = 1 \\[1em] \Rightarrow x - y = 1 \text{ ........(4)}\end{aligned}$$

On adding equations (3) and (4) we get,

⇒ (x + y) + (x - y) = 5 + 1

⇒ 2x = 6

⇒ x = $\frac{6}{2} = 3$.

Substituting value of x in equation (4) we get,

⇒ x - y = 1

⇒ 3 - y = 1

⇒ y = 3 - 1 = 2.

Hence, x = 3, y = 2.

$\frac{5}{x + 1} - \frac{2}{y - 1} = \frac{1}{2},\frac{10}{x + 1} + \frac{2}{y - 1} = \frac{5}{2}$

Substituting $\frac{1}{x + 1}= u, \frac{1}{y - 1} = v$ in $\frac{5}{x + 1} - \frac{2}{y - 1} = \frac{1}{2}$, we get:

⇒ 5u - 2v = $\frac{1}{2}$

⇒ 5u - 2v - $\frac{1}{2}$ = 0     ....(1)

Substituting $\frac{1}{x + 1}= u, \frac{1}{y - 1} = v$ in $\frac{10}{x + 1} + \frac{2}{y - 1} = \frac{5}{2}$, we get:

⇒ 10u + 2v = $\frac{5}{2}$

⇒ 10u + 2v - $\frac{5}{2}$ = 0     ....(2)

Multiply equation (1) and (2) by 2, we get,

⇒ $2\Big(5u - 2v - \frac{1}{2}\Big)$= 0

⇒ 10u - 4v - 1 = 0     .......(3)

⇒ $2\Big(10u + 2v - \frac{5}{2}\Big)$ = 0

⇒ 20u + 4v - 5 = 0     .........(4)

Applying cross-multiplication method for solving equations (3) and (4), we get :

$$\begin{aligned}\Rightarrow \frac{u}{(-4) \times (-5) - (4) \times (-1)} = \frac{v}{(-1) \times (20) - (-5) \times (10)} = \frac{1}{(10) \times (4) - (20) \times (-4)} \\[1em] \Rightarrow \frac{u}{(20) + 4} = \frac{v}{-20 + 50} = \frac{1}{40 + 80} \\[1em] \Rightarrow \frac{u}{24} = \frac{v}{30} = \frac{1}{120} \\[1em] \Rightarrow \frac{u}{24} = \frac{1}{120} \text{ and } \frac{v}{30} = \frac{1}{120} \\[1em] \Rightarrow u = \frac{24}{120} \text{ and } v = \frac{30}{120} \\[1em] \Rightarrow u = \frac{1}{5} \text{ and } v = \frac{1}{4}.\end{aligned}$$

Now we have u = $\frac{1}{5} \text{ and } v = \frac{1}{4}$,

$$\begin{aligned}\Rightarrow \frac{1}{x + 1} = u \\[1em] \Rightarrow \frac{1}{x + 1} = \frac{1}{5} \\[1em] \Rightarrow x + 1 = 5 \\[1em] \Rightarrow x = 5 - 1 = 4 \\[1em] \Rightarrow \frac{1}{y - 1} = v \\[1em] \Rightarrow \frac{1}{y - 1} = \frac{1}{4} \\[1em] \Rightarrow y - 1 = 4 \\[1em] \Rightarrow y = 4 + 1 = 5.\end{aligned}$$

Hence, x = 4, y = 5.

Let two whole numbers be x and y, where x > y.

Given,

Sum of numbers = 53

⇒ x + y = 53     .......(1)

Difference of numbers = 25

⇒ x - y = 25     ......(2)

Adding equations (1) and (2), we get :

⇒ (x + y) + (x - y) = 53 + 25

⇒ x + x + y - y = 78

⇒ 2x = 78

⇒ x = $\frac{78}{2}$ = 39.

Substituting value of x in equation (1), we get :

⇒ x + y = 53

⇒ 39 + y = 53

⇒ y = 53 - 39 = 14.

Hence, the numbers are 39 and 14.

Let the numbers be x and y, such that x < y.

Given,

The sum of two numbers exceeds thrice the smaller by 2.

∴ (x + y) - 3x = 2

⇒ x - 3x + y = 2

⇒ y - 2x = 2     ......(1)

Given,

Difference between the numbers = 19

⇒ y - x = 19     ......(2)

Subtracting equation (1) from (2), we get:

⇒ y - x - (y - 2x) = 19 - 2

⇒ y - x - y + 2x = 19 - 2

⇒ x = 17.

Substituting value of x in equation (1), we get :

⇒ y - 2x = 2

⇒ y - 2 × 17 = 2

⇒ y - 34 = 2

⇒ y = 36.

Hence, the numbers are 17 and 36.

Let two numbers be x and y, where x > y.

Given,

Sum of numbers = 51

⇒ x + y = 51

⇒ x = 51 - y     ......(1)

Given,

Difference of numbers when larger is doubled and the smaller is tripled = 12

⇒ 2x - 3y = 12     ......(2)

Substituting value of x from equation (1) in (2), we get :

⇒ 2(51 - y) - 3y = 12

⇒ 102 - 2y - 3y = 12

⇒ -5y = 12 - 102

⇒ -5y = -90

⇒ y = $\frac{90}{5} = 18$.

Substituting value of y in equation (1), we get :

⇒ x = 51 - y

⇒ x = 51 - 18

⇒ x = 33.

Hence, the numbers are 33 and 18.

Let two numbers be x and y.

Given,

Sum of twice the first and thrice the second = 103

⇒ 2x + 3y = 103

⇒ 2x = 103 - 3y

⇒ x = $\Big(\frac{103 - 3y}{2}\Big)$     ......(1)

Given,

Four times the first number exceeds seven times the second by 11.

⇒ 4x - 7y = 11

⇒ 4x = 7y + 11     .....(2)

Substituting value of x from equation (1) in (2), we get :

⇒ $4\Big(\frac{103 - 3y}{2}\Big)$ = 7y + 11

⇒ 2(103 - 3y) = 7y + 11

⇒ 206 - 6y = 7y + 11

⇒ 206 - 11 = 7y + 6y

⇒ 195 = 13y

⇒ y = $\frac{195}{13} = 15$.

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \Big(\frac{103 - 3y}{2}\Big) \\[1em] \Rightarrow x = \Big(\frac{103 - 3 \times 15}{2}\Big) \\[1em] \Rightarrow x = \Big(\frac{103 - 45}{2}\Big) \\[1em] \Rightarrow x = \Big(\frac{58}{2}\Big) = 29.\end{aligned}$$

Hence, the numbers are 29 and 15.

Let two numbers be x and y.

Given,

Sum of thrice the first and the second = 142

⇒ 3x + y = 142

⇒ y = 142 - 3x     .....(1)

Given,

Four times the first exceeds the second by 138.

⇒ 4x - y = 138

⇒ 4x = y + 138     ......(2)

Substituting value of y from equation (1) in (2), we get :

⇒ 4x = 142 - 3x + 138

⇒ 4x + 3x = 142 + 138

⇒ 7x = 280

⇒ x = $\frac{280}{7} = 40$.

Substituting value of x in equation (1), we get :

⇒ y = 142 - 3x

⇒ y = 142 - 3(40)

⇒ y = 142 - 120

⇒ y = 22.

Hence, the numbers are 40 and 22.

Let two numbers be x and y, where x > y.

Given,

4 times the smaller one is less than 3 times the larger one by 6.

⇒ 3x - 4y = 6

⇒ 4y = 3x - 6

⇒ y = $\Big(\frac{3x - 6}{4}\Big)$     .............(1)

Given,

Sum of the numbers is larger than 6 times their difference by 5.

⇒ x + y - 6(x - y) = 5

⇒ x + y - 6x + 6y = 5

⇒ x - 6x + y + 6y = 5

⇒ 7y - 5x = 5     ...........(2)

Substituting value of y from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow 7\Big(\frac{3x - 6}{4}\Big) - 5x = 5 \\[1em] \Rightarrow \Big(\frac{21x - 42}{4}\Big) - 5x = 5 \\[1em] \Rightarrow \Big(\frac{21x - 42 - 20x}{4}\Big) = 5 \\[1em] \Rightarrow \frac{x - 42}{4} = 5 \\[1em] \Rightarrow x - 42 = 20 \\[1em] \Rightarrow x = 20 + 42 \\[1em] \Rightarrow x = 62.\end{aligned}$$

Substituting value of x in equation (1), we get :

$$\begin{aligned}\Rightarrow y = \Big(\frac{3x - 6}{4}\Big) \\[1em] \Rightarrow y = \Big(\frac{3 \times 62 - 6}{4}\Big) \\[1em] \Rightarrow y = \Big(\frac{186 - 6}{4}\Big) \\[1em] \Rightarrow y = \Big(\frac{180}{4}\Big) = 45.\end{aligned}$$

Hence, the numbers are 62 and 45.

Let two numbers be x and y, where x > y.

Given,

If from twice the greater number of the two numbers, 45 is subtracted, the result is the other number.

⇒ 2x - 45 = y     .....(1)

Given,

If from twice the smaller number, 21 is subtracted, the result is the greater number.

⇒ 2y - 21 = x     .....(2)

Substituting value of y from equation (1) in (2), we get :

⇒ 2y - 21 = x

⇒ 2(2x - 45) - 21 = x

⇒ 4x - 90 - 21 = x

⇒ 4x - x - 111 = 0

⇒ 3x = 111

⇒ x = $\frac{111}{3} = 37.$

Substituting value of x in equation (1), we get :

⇒ 2x - 45 = y

⇒ 2(37) - 45 = y

⇒ 74 - 45 = y

⇒ y = 29.

Hence, the numbers are 37 and 29.

Let two numbers be x and y, where x > y.

Given,

Three times the larger divided by the smaller gives quotient 4.

⇒ 3x = 4y + 5

⇒ x = $\frac{4y + 5}{3}$     .....(1)

Given,

Six times the smaller divided by the larger gives quotient 4 and remainder 2.

⇒ 6y = 4x + 2     .....(2)

Substituting value of x from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow 6y = 4 \Big(\frac{4y + 5}{3}\Big) + 2 \\[1em] \Rightarrow 6y = \Big(\frac{16y + 20}{3}\Big) + 2 \\[1em] \Rightarrow 6y = \Big(\frac{16y + 20 + 6}{3}\Big) \\[1em] \Rightarrow 6y = \Big(\frac{16y + 26}{3}\Big) \\[1em] \Rightarrow 6y \times 3 = 16y + 26 \\[1em] \Rightarrow 18y = 16y + 26 \\[1em] \Rightarrow 18y - 16y = 26 \\[1em] \Rightarrow 2y = 26 \\[1em] \Rightarrow y = \frac{26}{2} \\[1em] \Rightarrow y = 13.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{4y + 5}{3} \\[1em] \Rightarrow x = \frac{4 \times 13 + 5}{3} \\[1em] \Rightarrow x = \frac{52 + 5}{3} \\[1em] \Rightarrow x = \frac{57}{3} = 19.\end{aligned}$$

Hence, the numbers are 19 and 13.

Let two numbers be x and y.

Given,

If 2 is added to each, the ratio becomes 1 : 2.

⇒ $\frac{x + 2}{y + 2} = \frac{1}{2}$

⇒ 2(x + 2) = y + 2

⇒ 2x + 4 = y + 2

⇒ y = 2x + 4 - 2

⇒ y = 2x + 2     ....(1)

Given,

If 4 is subtracted from each number, the ratio becomes 5 : 11.

⇒ $\frac{x - 4}{y - 4} = \frac{5}{11}$

⇒ 11(x - 4) = 5(y - 4)

⇒ 11x - 44 = 5y - 20

⇒ 11x - 5y = -20 + 44

⇒ 11x - 5y = 24     ....(2)

Substituting value of y from equation (1) in (2), we get :

⇒ 11x - 5(2x + 2) = 24

⇒ 11x - 10x - 10 = 24

⇒ x - 10 = 24

⇒ x = 24 + 10

⇒ x = 34.

Substituting value of x in equation (1), we get :

⇒ y = 2x + 2

⇒ y = 2 × 34 + 2

⇒ y = 68 + 2

⇒ y = 70.

Hence, the numbers are 34 and 70.

Let two numbers be x and y, where x > y.

Given,

Difference between two numbers = 12.

⇒ x - y = 12     .........(1)

Given,

Difference between their squares = 456

⇒ x² - y² = 456

By identity,

⇒ x² - y² = (x + y)(x - y)

⇒ (x + y)(x - y) = 456

Substituting value of x - y from equation (1) in above equation, we get :

⇒ (x + y)(12) = 456

⇒ (x + y) = $\frac{456}{12}$

⇒ x + y = 38     ........(2)

Adding equations (1) and (2),

⇒ x + y + x - y = 38 + 12

⇒ 2x = 50

⇒ x = $\frac{50}{2}$

⇒ x = 25.

Substituting value of x in equation (1), we get :

⇒ x - y = 12

⇒ 25 - y = 12

⇒ y = 25 - 12

⇒ y = 13.

Hence, the numbers are 25 and 13.

Let the numerator be x and denominator be y.

Thus, fraction = $\frac{x}{y}$

Given,

The fraction becomes $\frac{1}{2}$ when its numerator is increased by 6.

⇒ $\frac{x + 6}{y} = \frac{1}{2}$

⇒ 2(x + 6) = y

⇒ 2x + 12 = y     ..........(1)

Given,

The fraction becomes $\frac{1}{3}$ when its denominator is increased by 7.

⇒ $\frac{x}{y + 7} = \frac{1}{3}$

⇒ 3x = y + 7     .......(2)

Substituting value of y from equation (1) in (2), we get :

⇒ 3x = (2x + 12) + 7

⇒ 3x = 2x + 19

⇒ 3x - 2x = 19

⇒ x = 19.

Substituting value of x in equation (1), we get :

⇒ 2 × 19 + 12 = y

⇒ 38 + 12 = y

⇒ y = 50.

Hence, the fraction = $\frac{19}{50}$.

Let the numerator be x and denominator be y.

Thus, fraction = $\frac{x}{y}$

Given,

The fraction becomes $\frac{1}{2}$ when 1 is subtracted from the numerator and 1 is added to the denominator,

⇒ $\frac{x - 1}{y + 1} = \frac{1}{2}$

⇒ 2(x - 1) = y + 1

⇒ 2x - 2 = y + 1

⇒ y = 2x - 2 - 1

⇒ y = 2x - 3     .........(1)

Given,

The fraction becomes $\frac{1}{3}$ when 6 is subtracted from the numerator and 1 from the denominator,

⇒ $\frac{x - 6}{y - 1} = \frac{1}{3}$

⇒ 3(x - 6) = y - 1

⇒ 3x - 18 = y - 1

⇒ y = 3x - 18 + 1

⇒ y = 3x - 17     .........(2)

From equations (1) in (2), we get :

⇒ 3x - 17 = 2x - 3

⇒ 3x - 2x = -3 + 17

⇒ 3x - 2x = 14

⇒ x = 14.

Substituting value of x in equation (1), we get :

⇒ y = 2(14) - 3 = 28 - 3 = 25.

Fraction = $\frac{x}{y} = \frac{14}{25}$.

Hence, the fraction = $\frac{14}{25}$.

Let the numerator be x.

Given,

The denominator of a fraction is greater than its numerator by 9.

⇒ Denominator = x + 9

Thus, fraction = $\frac{x}{x + 9}$

Given,

The fraction becomes $\frac{2}{3}$, when 7 is subtracted from both numerator and denominator,

⇒ $\frac{x - 7}{(x + 9) - 7} = \frac{2}{3}$

⇒ $\frac{x - 7}{x + 2} = \frac{2}{3}$

⇒ 3(x - 7) = 2(x + 2)

⇒ 3x - 21 = 2x + 4

⇒ 3x - 2x = 4 + 21

⇒ x = 25.

The denominator is (x + 9) = 25 + 9 = 34.

Fraction = $\frac{x}{y} = \frac{25}{34}$.

Hence, the fraction = $\frac{25}{34}$.

Let the tens and unit digits of required number be x and y. x > y (according to hint)

Given,

Difference of digits of the number is 3.

⇒ x - y = 3

⇒ x = y + 3     .....(1)

Original number = 10x + y

Number obtained by reversing the digits = (10y + x)

Given,

4 times the number equals 7 times the number obtained by reversing its digits.

⇒ 4(10x + y) = 7(10y + x)

⇒ 40x + 4y = 70y + 7x

⇒ 40x - 7x + 4y - 70y = 0

⇒ 33x - 66y = 0     .....(2)

Substituting the value of x from equation (1) in (2), we get :

⇒ 33(y + 3) - 66y = 0

⇒ 33y + 33 × 3 - 66y = 0

⇒ 99 - 33y = 0

⇒ 99 - 33y = 0

⇒ 33y = 99

⇒ y = $\frac{99}{33}$

⇒ y = 3.

Substituting value of y in equation (1), we get :

⇒ x = y + 3

⇒ x = 3 + 3

⇒ x = 6.

Original number = 10x + y

= 10 × 6 + 3

= 63.

Hence, the number is 63.

According to question,

Original number is smaller than the number obtained by reversing its digits.

∴ In original number, unit's digit greater than ten's digit.

Let the ten's and unit's digit of required number be x and y respectively y > x.

Given,

⇒ y - x = 5

⇒ y = x + 5     .....(1)

Number obtained by reversing the digits = (10y + x)

Given,

8 times the number is equal to 3 times the number obtained by reversing the digits.

⇒ 8(10x + y) = 3(10y + x)

⇒ 80x + 8y = 30y + 3x

⇒ 80x - 3x + 8y - 30y = 0

⇒ 77x - 22y = 0     .....(2)

Substituting the value of x from equation (2) in 77x - 22y = 0, we get:

⇒ 77x - 22y = 0

⇒ 77x - 22 × (x + 5) = 0

⇒ 77x - 22x - 110 = 0

⇒ 55x - 110 = 0

⇒ 55x = 110

⇒ x = $\frac{110}{55}$

⇒ x = 2.

Substituting value of y in equation (2), we get :

⇒ y = x + 5

⇒ y = 2 + 5

⇒ y = 7.

Number = (10x + y) = 10 × 2 + 7 = 27.

Hence, the number is 27.

Let the ten's and unit's digits of required number be x and y respectively.

Given,

Sum of the digits of the number is 12.

⇒ x + y = 12

⇒ x = 12 - y     .....(1)

Original number = 10x + y

Number obtained by reversing the digits = 10y + x

Given,

On dividing the number by the number with its digits reversed, the result is $\Big(1\frac{3}{4}\Big)$.

$$\begin{aligned}\therefore \Big(\frac{10x + y}{10y + x}\Big) = \Big(1\frac{3}{4}\Big) \\[1em] \Rightarrow \Big(\frac{10x + y}{10y + x}\Big) = \frac{7}{4} \\[1em] \Rightarrow (10x + y) \times 4 = (10y + x) \times 7 \\[1em] \Rightarrow 40x + 4y = 70y + 7x \\[1em] \Rightarrow 40x - 7x + 4y - 70y = 0 \\[1em] \Rightarrow 33x - 66y = 0 \text{.....(2)}\end{aligned}$$

Substituting the value of x from equation (1) in (2), we get :

⇒ 33(12 - y) - 66y = 0

⇒ 396 - 33y - 66y = 0

⇒ 396 - 99y = 0

⇒ 99y = 396

⇒ y = $\frac{396}{99}$

⇒ y = 4.

Substituting value of y in equation (1), we get :

⇒ x = 12 - y

⇒ x = 12 - 4

⇒ x = 8.

Original number = (10x + y)

= 10 × 8 + 4

= 84.

Hence, the number is 84.

Let the ten's and unit's digit of required number be x and y respectively.

Number = 10x + y

Given,

On dividing the number by the sum of its digits, the quotient is 8.

$$\begin{aligned}\Rightarrow \Big(\frac{10x + y}{x + y}\Big) =8 \\[1em] \Rightarrow 10x + y = 8(x + y) \\[1em] \Rightarrow 10x + y = 8x + 8y \\[1em] \Rightarrow 10x - 8x + y - 8y = 0 \\[1em] \Rightarrow 2x - 7y = 0 \text{ .....(1)}\end{aligned}$$

Given,

On diminishing the ten’s digit by three times the unit’s digit, the remainder obtained is 1.

⇒ x - 3y = 1

⇒ x = 3y + 1     .........(2)

Substituting the value of x from equation (2) in (1), we get :

⇒ 2(3y + 1) - 7y = 0

⇒ 6y + 2 - 7y = 0

⇒ 2 - y = 0

⇒ y = 2.

Substituting value of y in equation (2), we get :

⇒ x = 3 × 2 + 1

⇒ x = 6 + 1

⇒ x = 7.

Number = 10x + y

= 10 × 7 + 2

= 72.

Hence, the number is 72.

Let the ten's and unit's digit of required number be x and y respectively.

Number = 10x + y

Given,

A number of two digits exceeds four times the sum of its digits by 6.

⇒ 10x + y - 4(x + y) = 6

⇒ 10x + y - 4x - 4y = 6

⇒ 6x - 3y = 6     .....(1)

Number obtained by reversing the digits = 10y + x

Given,

Number is increased by 9 on reversing its digits.

⇒ 10y + x = 10x + y + 9

⇒ 10y - y + x - 10x = 9

⇒ 9y - 9x = 9

⇒ y - x = 1

⇒ x = y - 1     .....(2)

Substituting the value of x from equation (2) in equation 1,

⇒ 6(y - 1) - 3y = 6

⇒ 6y - 6 - 3y = 6

⇒ 3y - 6 = 6

⇒ 3y = 6 + 6

⇒ 3y = 12

⇒ y = $\frac{12}{3}$

⇒ y = 4.

Substituting value of y in equation (2), we get :

⇒ x = 4 - 1

⇒ x = 3.

Number = 10x + y

= 10 × 3 + 4

= 34.

Hence, the number is 34.

Let the ten's and unit's digits of required number be x and y.

Number = 10x + y

Given,

Sum of digits = 12.

⇒ x + y = 12

⇒ x = 12 - y     .....(1)

Given,

Number obtained by reversing the digits = (10y + x)

If the digits are reversed, the new number is 12 less than twice the original number.

⇒ (10y + x) = 2(10x + y) - 12

⇒ 10y + x = 20x + 2y - 12

⇒ 10y - 2y + x - 20x = -12

⇒ 8y - 19x = - 12     .....(2)

Substituting the value of x from equation (1) in equation (2),

⇒ 8y - 19(12 - y) = -12

⇒ 8y - 228 + 19y = -12

⇒ 27y = -12 + 228

⇒ 27y = 216

⇒ y = $\frac{216}{27}$

⇒ y = 8.

Substituting value of y in equation (1), we get :

⇒ x = 12 - 8

⇒ x = 4.

The number is,

⇒ (10x + y) = 10 × 4 + 8 = 48.

Hence, the number is 48.

Let the cost of a pen be ₹ x and ₹ y be the cost of a pencil.

Given,

11 pens and 19 pencils together cost ₹ 502.

⇒ 11x + 19y = 502     ........(1)

Given,

19 pens and 11 pencils together cost ₹ 758.

⇒ 19x + 11y = 758     .........(2)

Subtracting equation (1) from equation (2), we get :

⇒ 19x + 11y - (11x + 19y) = 758 - 502

⇒ 19x - 11x + 11y - 19y = 256

⇒ 8x - 8y = 256

⇒ 8(x - y) = 256

⇒ x - y = $\frac{256}{8}$

⇒ x - y = 32

⇒ x = 32 + y .....(3)

Substituting value of x in equation (1), we get :

⇒ 11x + 19y = 502

⇒ 11(32 + y) + 19y = 502

⇒ 352 + 11y + 19y = 502

⇒ 30y = 502 - 352

⇒ 30y = 150

⇒ y = $\frac{150}{30}$

⇒ y = ₹ 5.

Substituting value of y in equation (3), we get :

⇒ x = 32 + y

⇒ x = 32 + 5

⇒ x = ₹ 37.

The amount to buy the 3 pens and 6 pencils is,

⇒ 3x + 6y

⇒ 3 × 37 + 6 × 5

⇒ 111 + 30

⇒ ₹ 141.

Hence, 3 pens and 6 pencils costs ₹ 141.

Let cost of sugar be ₹ x/kg and cost of rice be ₹ y/kg.

Given,

5 kg sugar and 7 kg rice together cost ₹ 258,

⇒ 5x + 7y = 258

⇒ 7y = 258 - 5x

⇒ y = $\frac{258 - 5x}{7}$     .....(1)

Given,

7 kg sugar and 5 kg rice together cost ₹ 246,

⇒ 7x + 5y = 246     .....(2)

Substituting value of y from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow 7x + 5\Big(\frac{258 - 5x}{7}\Big) = 246 \\[1em] \Rightarrow \Big(\frac{49x + 1290 - 25x}{7}\Big) = 246 \\[1em] \Rightarrow 24x + 1290 = 246 \times 7 \\[1em] \Rightarrow 24x + 1290 = 1722 \\[1em] \Rightarrow 24x = 1722 - 1290 \\[1em] \Rightarrow x = \frac{432}{24} = 18.\end{aligned}$$

Substituting value of x in equation (1), we get :

$$\begin{aligned}\Rightarrow y = \frac{258 - 5 \times 18}{7} \\[1em] \Rightarrow y = \frac{258 - 90}{7} \\[1em] \Rightarrow y = \frac{168}{7} \\[1em] \Rightarrow y = 24.\end{aligned}$$

Total cost of 8 kg sugar and 10 kg rice = 8x + 10y

= 8 × 18 + 10 × 24

= 144 + 240 = ₹ 384.

Hence, the total cost of 8 kg sugar and 10 kg rice = ₹ 384.

Let the present age of man be x years and his son be y years.

Given,

One year ago a man was four times as old as his son,

⇒ x - 1 = 4(y - 1)

⇒ x - 1 = 4y - 4

⇒ x = 4y - 4 + 1

⇒ x = 4y - 3     ....(1)

Given,

After 6 years, the man's age will exceed twice his son's age by 9,

⇒ x + 6 = 2(y + 6) + 9

⇒ x + 6 = 2y + 12 + 9

⇒ x = 2y + 12 + 9 - 6

⇒ x = 2y + 15     .....(2)

From equation (1) in (2), we get :

⇒ 4y - 3 = 2y + 15

⇒ 4y - 2y = 15 + 3

⇒ 2y = 18

⇒ y = $\frac{18}{2}$ = 9.

Substituting value of y in equation (1), we get :

⇒ x = 4y - 3

⇒ x = 4(9) - 3

⇒ x = 36 - 3 = 33.

Hence, man's present age = 33 years and son's present age = 9 years.

Let x be the present age of A and y be the present age of B.

Given,

5 years ago, A was thrice as old as B.

⇒ x - 5 = 3(y - 5)

⇒ x - 5 = 3y - 15

⇒ x = 3y - 15 + 5

⇒ x = 3y - 10     ...,,,,.(1)

Given,

10 years later, A shall be twice as old as B.

⇒ x + 10 = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x = 2y + 20 - 10

⇒ x = 2y + 10     .........(2)

From equation (1) and (2), we get :

⇒ 3y - 10 = 2y + 10

⇒ 3y - 2y = 10 + 10

⇒ y = 20.

Substituting value of y in equation (1), we get :

⇒ x = 3y - 10

⇒ x = 3(20) - 10

⇒ x = 60 - 10 = 50.

Hence, A's present age = 50 years and B's present age = 20 years.

Let monthly income of A be 7x and B be 5x.

Let monthly expenditure of A be 3y and B be 2y.

Given,

Both A and B save ₹ 1500 per month.

For A :

⇒ 7x - 3y = 1500

⇒ 7x = 1500 + 3y

⇒ x = $\frac{1500 + 3y}{7}$     .....(1)

For B :

⇒ 5x - 2y = 1500     ......(2)

Substituting value of x from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow 5x - 2y = 1500 \\[1em] \Rightarrow 5\Big(\frac{1500 + 3y}{7}\Big) - 2y = 1500 \\[1em] \Rightarrow \Big(\frac{7500 + 15y}{7}\Big) - 2y = 1500 \\[1em] \Rightarrow \Big(\frac{7500 + 15y - 14y}{7}\Big) = 1500 \\[1em] \Rightarrow 7500 + y = 10500 \\[1em] \Rightarrow y = 10500 - 7500 \\[1em] \Rightarrow y = 3000.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{1500 + 3y}{7} \\[1em] \Rightarrow x = \frac{1500 + 3 \times 3000}{7} \\[1em] \Rightarrow x = \frac{1500 + 9000}{7} \\[1em] \Rightarrow x = \frac{10500}{7} = 1500.\end{aligned}$$

A's income = ₹ 7x = ₹ 7 × 1500 = ₹ 10,500,

B's income = ₹ 5x = ₹ 5 × 1500 = ₹ 7,500.

Hence, A's income = ₹ 10,500, B's income = ₹ 7,500.

Let x litres be the quantity of the 90% acid solution and y litres be the quantity of the 97% acid solution.

Given,

Total volume = 21 litres

⇒ x + y = 21

⇒ x = 21 - y     ....(1)

Als,

⇒ 90% of x + 97% of y = 95% of 21

⇒ 0.90x + 0.97y = 0.95 × 21

⇒ 0.90x + 0.97y = 19.95     ....(2)

Substituting value of x from equation (1) in (2), we get :

⇒ 0.90(21 - y) + 0.97y = 19.95

⇒ 18.9 - 0.90y + 0.97y = 19.95

⇒ 18.9 + 0.07y = 19.95

⇒ 0.07y = 19.95 - 18.9

⇒ 0.07y = 1.05

⇒ y = $\frac{1.05}{0.07} = 15$.

Substituting value of y in equation (1), we get :

⇒ x = 21 - y

⇒ x = 21 - 15

⇒ x = 6.

Hence, 6 litres of 90% acid solution and 15 litres of 97% acid solution are mixed.

Let x and y be the initial number of pupils in examination hall A and B respectively.

Given,

If 12 pupils are sent from A to B, the number of pupils in each room becomes the same.

⇒ x - 12 = y + 12

⇒ x = y + 12 + 12

⇒ x = 24 + y     .....(1)

Given,

If 11 pupils are sent from B to A, then the number of pupils in A is double their number in B.

⇒ x + 11 = 2(y - 11)

⇒ x + 11 = 2y - 22

⇒ x - 2y = -22 - 11

⇒ x - 2y = -33     ........(2)

Substituting value of x from equation (1) in (2), we get :

⇒ 24 + y - 2y = -33

⇒ 24 - y = -33

⇒ y = 24 + 33

⇒ y = 57.

Substituting value of y in equation (1), we get :

⇒ x = 24 + y

⇒ x = 24 + 57

⇒ x = 81.

Hence, no. of pupils in examination hall A = 81 and examination hall B = 57.

Let the number of marbles A has be x, and the number of marbles B has be y.

After B gives 30 marbles to A,

⇒ x + 30 = 2(y - 30)

⇒ x + 30 = 2y - 60

⇒ x = 2y - 60 - 30

⇒ x = 2y - 90     .......(1)

After A gives 10 marbles to B,

⇒ y + 10 = 3(x - 10)

⇒ y + 10 = 3x - 30

⇒ y = 3x - 30 - 10

⇒ y = 3x - 40     ........(2)

Substituting value of x from equation (1) in (2), we get :

⇒ y = 3(2y - 90) - 40

⇒ y = 6y - 270 - 40

⇒ y = 6y - 310

⇒ y - 6y = -310

⇒ -5y = -310

⇒ y = $\frac{-310}{-5} = 62$.

Substituting value of y in equation (1), we get :

⇒ x = 2y - 90

⇒ x = 2(62) - 90

⇒ x = 124 - 90

⇒ x = 34.

Hence, A has 34 marbles and B has 62 marbles.

Let x be present age of the man and y be the present age of son.

Given,

The present age of a man is 3 years more than three times the age of his son,

⇒ x = 3y + 3     ........(1)

Given,

Three years hence, the man’s age will be 10 years more than twice the age of his son.

⇒ x + 3 = 2(y + 3) + 10

⇒ x + 3 = 2y + 6 + 10

⇒ x = 2y + 16 - 3

⇒ x = 2y + 13     ........(2)

From equation (1) and (2), we get :

⇒ 3y + 3 = 2y + 13

⇒ 3y - 2y = 13 - 3

⇒ y = 10.

Substituting value of y in equation (1), we get :

⇒ x = 3y + 3

⇒ x = 3(10) + 3

⇒ x = 30 + 3

⇒ x = 33.

Hence, the present age of son = 10 years and that of man = 33 years.

Let x meters be the length of a room and y meters be the breadth of the room.

Given,

The length of a room exceeds its breadth by 3 m,

⇒ x = y + 3     ....(1)

Given,

If length is increased by 3 and breadth decreased by 2, area remains the same,

⇒ (x + 3)(y - 2) = xy

⇒ (xy - 2x + 3y - 6) = xy

⇒ xy - xy = 2x - 3y + 6

⇒ 2x - 3y + 6 = 0     ....(2)

Substituting value of x from equation (1) in (2), we get :

⇒ 2(y + 3) - 3y + 6 = 0

⇒ 2y + 6 - 3y + 6 = 0

⇒ -y + 12 = 0

⇒ y = 12.

Substituting value of y in equation (1), we get :

⇒ x = y + 3

⇒ x = 12 + 3

⇒ x = 15.

Hence, length and breadth of the room are 15 m and 12 m respectively.

Let x meters be the length and y meters be the breadth of the rectangle.

Given,

If the length is reduced by 5 m and breadth is increased by 3 m, then the area reduces by 8 m².

⇒ (x - 5)(y + 3) = xy - 8

⇒ (xy + 3x - 5y - 15) = xy - 8

⇒ 3x - 5y - 15 + 8 = xy - xy

⇒ 3x - 5y - 7 = 0

⇒ 3x = 5y + 7

⇒ x = $\frac{5y + 7}{3}$     ........(1)

Given,

If the length is increased by 3 m and breadth by 2 m, then the area increases by 74 m².

⇒ (x + 3)(y + 2) = xy + 74

⇒ (xy + 2x + 3y + 6) = xy + 74

⇒ 2x + 3y + 6 - 74 = xy - xy

⇒ 2x + 3y - 68 = 0     .......(2)

Substituting value of x from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow 2 \Big(\frac{5y + 7}{3}\Big) + 3y - 68 = 0 \\[1em] \Rightarrow \Big(\frac{10y + 14}{3}\Big) + 3y = 68 \\[1em] \Rightarrow \Big(\frac{10y + 14 + 9y}{3}\Big) = 68 \\[1em] \Rightarrow 19y + 14 = 68 \times 3 \\[1em] \Rightarrow 19y + 14 = 204 \\[1em] \Rightarrow 19y = 204 - 14 \\[1em] \Rightarrow 19y = 190 \\[1em] \Rightarrow y = \frac{190}{19} = 10.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{5y + 7}{3} \\[1em] \Rightarrow x = \frac{5 \times 10 + 7}{3} \\[1em] \Rightarrow x = \frac{57}{3} = 19.\end{aligned}$$

Hence, the length and breadth of rectangles are 19 m and 10 m respectively.

Let x km/hr be the speed of motorboat in still water and y km/hr be the speed of stream.

Downstream speed = x + y km/h

Upstream speed = x - y km/h

Given,

Time $= \frac{\text{Distance}}{\text{Speed}}$

Motorboat takes 6 hours to cover 100 km downstream and 30 km upstream.

$\Rightarrow \frac{100}{x + y} + \frac{30}{x - y} = 6$     .........(1)

Given,

Motorboat goes 75 km downstream and returns back to its starting point in 8 hours.

$\Rightarrow \frac{75}{x + y} + \frac{75}{x - y} = 8$     ....(2)

Substituting, u = $\frac{1}{x + y}$, v = $\frac{1}{x - y}$ in equation (1), we get :

⇒ 100u + 30v = 6

⇒ 10(10u + 3v) = 6

⇒ (10u + 3v) = $\frac{6}{10}$     ........(3)

Substituting, u = $\frac{1}{x + y}$, v = $\frac{1}{x - y}$ in equation (2), we get :

⇒ 75u + 75v = 8

⇒ 75(u + v) = 8

⇒ u + v = $\frac{8}{75}$

⇒ u = $\frac{8}{75} - v$     ........(4)

Substituting value of u from equation (4) in (3), we get :

$$\begin{aligned}\Rightarrow 10 \Big(\frac{8}{75} - v\Big) + 3v = \frac{6}{10} \\[1em] \Rightarrow \frac{80}{75} - 10v + 3v = \frac{6}{10} \\[1em] \Rightarrow \frac{80}{75} - 7v = \frac{6}{10} \\[1em] \Rightarrow 7v = \frac{80}{75} - \frac{6}{10} \\[1em] \Rightarrow 7v = \frac{160 - 90}{150} \\[1em] \Rightarrow 7v = \frac{70}{150} \\[1em] \Rightarrow v = \frac{7}{15 \times 7} \\[1em] \Rightarrow v = \frac{1}{15}.\end{aligned}$$

Substituting value of v in equation (4), we get :

$$\begin{aligned}\Rightarrow u = \frac{8}{75} - v \\[1em] \Rightarrow u = \frac{8}{75} - \frac{1}{15} \\[1em] \Rightarrow u = \frac{8 - 5}{75} \\[1em] \Rightarrow u = \frac{3}{75} \\[1em] \Rightarrow u = \frac{1}{25}.\end{aligned}$$

Since,

$$\begin{aligned}\Rightarrow u = \frac{1}{x + y} \\[1em] \Rightarrow \frac{1}{25} = \frac{1}{x + y}\end{aligned}$$

⇒ x + y = 25     .......(5)

$$\begin{aligned}\Rightarrow v = \frac{1}{x - y} \\[1em] \Rightarrow \frac{1}{15} = \frac{1}{x - y}\end{aligned}$$

⇒ x - y = 15     ........(6)

Adding equations (5) and (6),

⇒ x + y + x - y = 25 + 15

⇒ 2x = 40

⇒ x = $\frac{40}{2} = 20$.

Substituting value of x in equation (6), we get :

⇒ x - y = 15

⇒ 20 - y = 15

⇒ 20 - 15 = y

⇒ y = 5.

Hence, the speed of the motorboat in still water = 20 km/hr and the speed of the stream = 5 km/hr.

Let x be the cost of chair and y be the cost of table.

Given,

When sold for ₹ 2,178, he makes profit of 12% on the chair and 16% on the table.

$$\begin{aligned}\therefore x + \frac{12}{100}x + y + \frac{16}{100}y = 2178 \\[1em] \Rightarrow \frac{100x + 12x}{100} + \frac{100y + 16y}{100} = 2178 \\[1em] \Rightarrow \frac{112x}{100} + \frac{116y}{100} = 2178 \\[1em] \Rightarrow 112x + 116y = 2178 \times 100 \\[1em] \Rightarrow 112x + 116y = 217800\\[1em] \Rightarrow 4(28x + 29y) = 217800 \\[1em] \Rightarrow 28x + 29y = \frac{217800}{4} \\[1em] \Rightarrow 28x + 29y = 54450 \text{ ....(1) }\end{aligned}$$

Given,

When sold for ₹ 2,154, he makes profit of 16% on the chair and 12% on the table.

$$\begin{aligned}\therefore x + \frac{16}{100}x + y + \frac{12}{100}y = 2154 \\[1em] \Rightarrow \frac{100x + 16x}{100} + \frac{100y + 12}{100} = 2154 \\[1em] \Rightarrow \frac{116x}{100} + \frac{112y}{100} = 2154 \\[1em] \Rightarrow 116x + 112y = 2154 \times 100 \\[1em] \Rightarrow 116x + 112y = 215400 \\[1em] \Rightarrow 4(29x + 28y) = 215400 \\[1em] \Rightarrow 29x + 28y = \frac{215400}{4} \\[1em] \Rightarrow 29x + 28y = 53850 \text{ ........(2) }\end{aligned}$$

Subtracting equation eqn 1 from 2, we get :

⇒ 29x + 28y - (28x + 29y) = 53850 - 54450

⇒ 29x + 28y - 28x - 29y = -600

⇒ x - y = -600

⇒ x = y - 600     ....(3)

Substituting value of x from equation (3) in equation (1), we get :

⇒ 28x + 29y = 54450

⇒ 28(y - 600) + 29y = 54450

⇒ 28y - 16800 + 29y = 54450

⇒ 57y = 54450 + 16800

⇒ 57y = 71250

⇒ y = $\frac{71250}{57}$

⇒ y = ₹ 1,250

Substituting value of y in equation (2), we get :

⇒ 29x + 28 × 1250 = 53850

⇒ 29x + 35000 = 53850

⇒ 29x = 53850 - 35000

⇒ 29x = 18850

⇒ x = $\frac{18850}{29}$

⇒ x = ₹ 650

Hence, cost price of table = ₹ 1,250 and cost price of chair = ₹ 650.

Let x km/hr be the speed of train and y km/hr be the speed of car.

$Time = \frac{\text{Distance}}{\text{Speed}}$

Given,

120 km by train, rest (600 - 120 = 480 km) by car takes 8 hours.

$\Rightarrow \frac{120}{x} + \frac{480}{y} = 8$ ........(1)

Given,

200 km by train, rest (600 - 200 = 400 km) by car takes 8 hours 20 mins = $8 + \frac{20}{60} = 8 + \frac{1}{3} = \frac{25}{3}$ hours.

$\Rightarrow \frac{200}{x} + \frac{400}{y} = \frac{25}{3}$ ..........(2)

Substituting, u = $\frac{1}{x}$, v = $\frac{1}{y}$ in equation (1), we get :

⇒ 120u + 480v = 8

⇒ 120u + 480v - 8 = 0

⇒ 8(15u + 60v - 1) = 0

⇒ 15u + 60v - 1 = 0

⇒ 15u = 1 - 60v

⇒ u = $\frac{1 - 60v}{15}$     ..........(3)

Substituting, u = $\frac{1}{x}$, v = $\frac{1}{y}$ in equation (2), we get :

⇒ 200u + 400v = $\frac{25}{3}$     .........(4)

Substituting value of u from equation (3) in equation (4), we get :

$$\begin{aligned}\Rightarrow 200 \Big(\frac{1 - 60v}{15}\Big) + 400v = \frac{25}{3} \\[1em] \Rightarrow \Big(\frac{200 - 12000v}{15}\Big) + 400v = \frac{25}{3} \\[1em] \Rightarrow \Big(\frac{200 - 12000v + 6000v}{15}\Big) = \frac{25}{3} \\[1em] \Rightarrow 200 - 6000v = \frac{25}{3} \times 15 \\[1em] \Rightarrow 200 - 6000v = 25 \times 5 \\[1em] \Rightarrow 200 - 6000v = 125 \\[1em] \Rightarrow 6000v = 200 - 125 \\[1em] \Rightarrow 6000v = 75 \\[1em] \Rightarrow v = \frac{75}{6000} \\[1em] \Rightarrow v = \frac{1}{80}.\end{aligned}$$

Substituting value of v in equation (1), we get :

$$\begin{aligned}\Rightarrow u = \frac{1 - 60 \times \frac{1}{80}}{15} \\[1em] \Rightarrow u = \frac{1 - \frac{3}{4}}{15} \\[1em] \Rightarrow u = \frac{\frac{1}{4}}{15} \\[1em] \Rightarrow u = \frac{1}{60}.\end{aligned}$$

Substituting,

⇒ u = $\frac{1}{x}$

⇒ $\frac{1}{60} = \frac{1}{x}$

⇒ x = 60.

⇒ v = $\frac{1}{y}$

⇒ $\frac{1}{80} = \frac{1}{y}$

⇒ y = 80.

Hence, speed of train = 60 km/h, speed of car = 80 km/h.

Lets assume that one man takes x days to do work and y days for one boy.

So, the amount of work done by 1 man in 1 day = $\frac{1}{x}$

So, the amount of work done by 1 boy in 1 day = $\frac{1}{y}$

Given,

6 men and 8 boys finish the work in 14 days,

$\Rightarrow \frac{6}{x} + \frac{8}{y} = \frac{1}{14} \text{ ....(1) }$

Given,

8 men and 12 boys finish the same work in 10 days,

$\Rightarrow \frac{8}{x} + \frac{12}{y} = \frac{1}{10}$     ....(2)

Multiply equation (1) by 2, we get:

$$\begin{aligned}\Rightarrow 2 \Big(\frac{6}{x} + \frac{8}{y}\Big) = \frac{1}{14} \times 2 \\[1em] \Rightarrow \frac{12}{x} + \frac{16}{y} = \frac{1}{7} \text{ ....(3) }\end{aligned}$$

Multiply equation (2) by $\frac{3}{2}$, we get:

$$\begin{aligned}\Rightarrow \frac{3}{2} \Big(\frac{8}{x} + \frac{12}{y}\Big) = \frac{1}{10} \times \frac{3}{2} \\[1em] \Rightarrow \frac{12}{x} + \frac{18}{y}= \frac{3}{20} \text{ ....(4) }\end{aligned}$$

Subtracting equation (3) from (4), we get:

$$\begin{aligned}\Rightarrow \Big(\frac{12}{x} + \frac{18}{y}\Big) - \Big(\frac{12}{x} + \frac{16}{y}\Big) = \frac{3}{20} - \frac{1}{7} \\[1em] \Rightarrow \Big(\frac{12}{x} + \frac{18}{y} - \frac{12}{x} - \frac{16}{y}\Big) = \frac{21}{140} - \frac{20}{140} \\[1em] \Rightarrow \frac{18}{y} - \frac{16}{y} = \frac{1}{140} \\[1em] \Rightarrow \frac{2}{y} = \frac{1}{140} \\[1em] \Rightarrow y = 2 \times 140 = 280.\end{aligned}$$

Substituting value of y in equation (1), we get:

$$\begin{aligned}\Rightarrow \frac{6}{x} + \frac{8}{y} = \frac{1}{14} \\[1em] \Rightarrow \frac{6}{x} + \frac{8}{280} = \frac{1}{14} \\[1em] \Rightarrow \frac{6}{x} = \frac{1}{14} - \frac{8}{280} \\[1em] \Rightarrow \frac{6}{x} = \frac{20}{280} - \frac{8}{280} \\[1em] \Rightarrow \frac{6}{x} = \frac{12}{280} \\[1em] \Rightarrow \frac{6 \times 280}{12} = x \\[1em] \Rightarrow x = 140.\end{aligned}$$

Hence, one man can finish the work in = 140 days and one boy finish the work in = 280 days .

Let Number of 25-P coins be x and number of 50-P coins be y.

Given,

Total number of coins = 80,

⇒ x + y = 80

⇒ x = 80 - y     .........(1)

Given,

Total value = ₹ 25 = 2500 paise,

⇒ 25x + 50y = 2500     .......(2)

Substituting value of x from equation (1) in 25x + 50y = 2500, we get :

⇒ 25x + 50y = 2500

⇒ 25(80 - y) + 50y = 2500

⇒ 2000 - 25y + 50y = 2500

⇒ -25y + 50y = 2500 - 2000

⇒ 25y = 500

⇒ y = $\frac{500}{25}$

⇒ y = 20.

Substituting value of y in equation (1), we get :

⇒ x = 80 - y

⇒ x = 80 - 20

⇒ x = 60.

Hence, number of 25-P coins = 60 and number of 50-P coins = 20.

Let A's one day work be x and B's one day work be y.

According to first condition given in the problem,

A works $1\frac{1}{2}$ times of B,

$\Rightarrow x = 1\frac{1}{2}y$

$\Rightarrow x = \frac{3}{2}y$

⇒ 2x = 3y

⇒ 2x - 3y = 0     ......(1)

Also given, A and B together can do a piece of work in 6 days.

$\therefore x + y = \frac{1}{6}$

⇒ 6(x + y) = 1

⇒ 6x + 6y = 1     ...(2)

Multiplying (1) by 2 we get,

⇒ 2(2x - 3y) = 2 × 0

⇒ 4x - 6y = 0     ...(3)

Adding equations (2) and (3) we get,

⇒ 6x + 6y + 4x - 6y = 1 + 0

⇒ 10x = 1

⇒ x = $\frac{1}{10}$

Substituting value of x in equation (1), we get :

⇒ 2 × $\frac{1}{10}$ - 3y = 0

⇒ $\frac{1}{5}$ - 3y = 0

⇒ 3y = $\frac{1}{5}$

⇒ y = $\frac{1}{15}$

Since, A's one day work is x and B's one day work is y, so A can do complete work in $\frac{1}{x}$ and B can do work in $\frac{1}{y}$ days.

$\frac{1}{x} = \frac{1}{\frac{1}{10}}$ = 10 days

$\frac{1}{y} = \frac{1}{\frac{1}{15}}$ = 15 days

Hence, A can finish the work in 10 days while B can finish the work in 15 days.

Given,

Equations: 3x - 2y = 5, x + 2y = -1

⇒ 3x - 2y = 5

⇒ 3x - 5 = 2y

⇒ y = $\frac{3x - 5}{2}$     ....(1)

Substituting value of y from equation (1) in x + 2y = -1, we get :

⇒ x + $2\Big(\frac{3x - 5}{2}\Big)$ = -1

⇒ x + 3x - 5 = -1

⇒ 4x = -1 + 5

⇒ 4x = 4

⇒ x = $\frac{4}{4}$

⇒ x = 1.

Substituting value of x in equation (1), we get :

$$\begin{aligned}\Rightarrow y = \frac{3x - 5}{2} \\[1em] \Rightarrow y = \frac{3 \times 1 - 5}{2} \\[1em] \Rightarrow y = \frac{-2}{2} \\[1em] \Rightarrow y = -1.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Equations: $\frac{x}{2} - \frac{y}{3} = 0, \frac{3x}{2} + \frac{2y}{3} + 10 = 0$

Solving equation $\frac{x}{2} - \frac{y}{3} = 0$,

$$\begin{aligned}\Rightarrow \frac{3x - 2y}{6} = 0 \\[1em] \Rightarrow 3x - 2y = 0 \times 6 \\[1em] \Rightarrow 3x = 2y \\[1em] \Rightarrow x = \frac{2y}{3} \text{ ....(1)}\end{aligned}$$

$\frac{3x}{2} + \frac{2y}{3} + 10 = 0 \text{ ....(2)}$

Substituting value of x from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow \frac{3\Big(\frac{2y}{3}\Big)}{2} + \frac{2y}{3} + 10 = 0 \\[1em] \Rightarrow \frac{2y}{2} + \frac{2y}{3} + 10 = 0 \\[1em] \Rightarrow \frac{6y + 4y + 60}{6} = 0 \\[1em] \Rightarrow 10y + 60 = 0 \\[1em] \Rightarrow 10y = -60 \\[1em] \Rightarrow y = \frac{-60}{10} = -6.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{2y}{3} \\[1em] \Rightarrow x = \frac{2 \times (-6)}{3} \\[1em] \Rightarrow x = \frac{-12}{3} = -4.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

Equations: $2x + \frac{1}{y} = 0$ and $3x + \frac{1}{2y} = -2$

Solving equation 1,

⇒ $2x + \frac{1}{y} = 0$

⇒ 2x = $-\frac{1}{y}$

⇒ x = $\frac{-1}{2y}$     .......(1)

Substituting value of x from equation (1) in $3x + \frac{1}{2y} = -2$, we get :

$$\begin{aligned}\Rightarrow 3\Big(\frac{-1}{2y}\Big) + \frac{1}{2y} = -2 \\[1em] \Rightarrow \frac{-3}{2y} + \frac{1}{2y} = -2 \\[1em] \Rightarrow \frac{-2}{2y} = -2 \\[1em] \Rightarrow y = \frac{-2}{-2 \times 2} \\[1em] \Rightarrow y = \frac{1}{2}.\end{aligned}$$

Substituting y = $\frac{1}{2}$ in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{-1}{2y} \\[1em] \Rightarrow x = \frac{-1}{2\Big(\frac{1}{2}\Big)} \\[1em] \Rightarrow x = -1.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Equations: $\frac{x}{6} + 6 = y, \frac{3x}{4} = 1 + y$

Solving first equation,

⇒ $\frac{x}{6} + 6 = y$

⇒ $\frac{x + 36}{6} = y$

⇒ x + 36 = 6y

⇒ x = 6y - 36     .......(1)

Substituting value of x from equation (1) in $\frac{3x}{4} = 1 + y$, we get :

$$\begin{aligned}\Rightarrow \frac{3(6y - 36)}{4} = 1 + y \\[1em] \Rightarrow 18y - 108 = 4(1 + y) \\[1em] \Rightarrow 18y - 108 = 4 + 4y \\[1em] \Rightarrow 18y - 4y = 4 + 108 \\[1em] \Rightarrow 14y = 112 \\[1em] \Rightarrow y = \frac{112}{14} = 8.\end{aligned}$$

Substituting value of y in equation (1), we get :

⇒ x = 6y - 36

⇒ x = 6(8) - 36

⇒ x = 48 - 36 = 12.

Hence, option 1 is the correct option.

Given,

⇒ (2x - 1)² + (3y - 1)² = 0

⇒ (2x - 1)² = 0 and (3y - 1)² = 0

⇒ (2x - 1) = 0 and 3y - 1 = 0

⇒ 2x = 1 and 3y = 1

⇒ x = $\frac{1}{2}$ and y = $\frac{1}{3}$.

Substituting value of x and y in $\Big(\frac{1}{x^2} + \frac{1}{y^2}\Big)$, we get :

$$\begin{aligned}\Rightarrow \Big(\frac{1}{x^2} + \frac{1}{y^2}\Big) \\[1em] \Rightarrow \frac{1}{\Big(\frac{1}{2}\Big)^2} + \frac{1}{\Big(\frac{1}{3}\Big)^2} \\[1em] \Rightarrow \frac{1}{\frac{1}{4}} + \frac{1}{\frac{1}{9}} \\[1em] \Rightarrow 4 + 9 \\[1em] \Rightarrow 13.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Equations: $\sqrt{5}x - \sqrt{7}y = 0, \sqrt{3}y + \sqrt{13}x = 0$

Solving first equation,

$$\begin{aligned}\Rightarrow \sqrt{5}x - \sqrt{7}y = 0 \\[1em] \Rightarrow \sqrt{5}x = \sqrt{7}y \\[1em] \Rightarrow x = \frac{\sqrt{7}}{\sqrt{5}}y \text{ ....(1)}\end{aligned}$$

Substituting value of x from equation (1) in $\sqrt{3}y + \sqrt{13}x = 0$, we get :

$$\begin{aligned}\Rightarrow \sqrt{3}y + \sqrt{13}x = 0 \\[1em] \Rightarrow \sqrt{3}y + \sqrt{13} \times \frac{\sqrt{7}}{\sqrt{5}}y = 0 \\[1em] \Rightarrow y \Big(\sqrt{3} + \frac{\sqrt{91}}{\sqrt{5}}\Big) = 0.\end{aligned}$$

Since, $\Big(\sqrt{3} + \frac{\sqrt{91}}{\sqrt{5}}\Big)$ is not equal to zero thus, y = 0.

Substituting value of y = 0 in equation (1), we get :

$\Rightarrow x = \frac{\sqrt{7}}{\sqrt{5}}y = \frac{\sqrt{7}}{\sqrt{5}} \times 0$ = 0.

Hence, option 1 is the correct option.

Given,

Equations: 0.4x + 3y = 1.2 and 7x - 2y = $\frac{17}{6}$

Solving first equation,

⇒ 0.4x + 3y = 1.2

Multiplying both sides of the equation by 10,

⇒ 10(0.4x + 3y) = 10 × 1.2

⇒ 4x + 30y = 12

⇒ 4x = 12 - 30y

⇒ x = $\frac{12 - 30y}{4}$     .......(1)

⇒ 7x - 2y = $\frac{17}{6}$     .......(2)

Substituting value of x from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow 7 \Big(\frac{12 - 30y}{4}\Big) - 2y = \frac{17}{6} \\[1em] \Rightarrow \frac{84 - 210y}{4} - 2y = \frac{17}{6} \\[1em] \Rightarrow \frac{84 - 210y - 8y}{4} = \frac{17}{6} \\[1em] \Rightarrow 84 - 218y = 4 \times \frac{17}{6} \\[1em] \Rightarrow 84 - 218y = 2 \times \frac{17}{3} \\[1em] \Rightarrow 3(84 - 218y) = 17 \times 2 \\[1em] \Rightarrow 252 - 654y = 34 \\[1em] \Rightarrow 654y = 252 - 34 \\[1em] \Rightarrow 654y = 218 \\[1em] \Rightarrow y = \frac{218}{654} \\[1em] \Rightarrow y = \frac{1}{3}.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{12 - 30y}{4} \\[1em] \Rightarrow x = \frac{12 - 30 \times \frac{1}{3}}{4} \\[1em] \Rightarrow x = \frac{12 - 10}{4} \\[1em] \Rightarrow x = \frac{2}{4} = \frac{1}{2}.\end{aligned}$$

Hence, option 2 is the correct option.

Let ₹ x be the cost of table and ₹ y be cost of the chair.

Given,

2 tables and 3 chairs together cost ₹ 1,075.

⇒ 2x + 3y = 1075     .......(1)

3 tables and 8 chairs together cost ₹ 1,875.

⇒ 3x + 8y = 1875     .......(2)

Multiplying equation (1) by 3, we get :

⇒ 3(2x + 3y) = 1075 × 3

⇒ 6x + 9y = 3225     ......(3)

Multiplying equation (2) by 2,

⇒ 2(3x + 8y) = 1875 × 2

⇒ 6x + 16y = 3750     ....(4)

Subtracting equation (3) from (4) we get,

⇒ (6x + 16y) - (6x + 9y) = 3750 - 3225

⇒ 6x + 16y - 6x - 9y = 525

⇒ 7y = 525

⇒ y = $\frac{525}{7}$ = ₹ 75.

Substituting value of y in equation (1), we get :

⇒ 2x + 3(75) = 1075

⇒ 2x + 225 = 1075

⇒ 2x = 1075 - 225

⇒ 2x = 850

⇒ x = $\frac{850}{2}$ = ₹ 425.

The cost of 4 tables and 5 chairs,

⇒ 4x + 5y = 4 × 425 + 5 × 75

= 1700 + 375 = ₹ 2,075.

Hence, option 3 is the correct option.

Let digit at ten's place be x and unit's place be y.

Number = 10x + y,

Given,

Sum of the digits of a two-digit = 9

⇒ x + y = 9     ....(1)

Given,

Nine times the number is twice the number obtained by reversing the order of the digits,

⇒ 9(10x + y) = 2(10y + x)

⇒ 90x + 9y = 20y + 2x

⇒ 90x - 2x = 20y - 9y

⇒ 88x = 11y

⇒ y = $\frac{88}{11}x$

⇒ y = 8x     ....(2)

Substituting value of y from equation (2) in equation (1), we get :

⇒ x + 8x = 9

⇒ 9x = 9

⇒ x = $\frac{9}{9}$ = 1.

Substituting value of x in equation (2), we get :

⇒ y = 8x

⇒ y = 8(1) = 8.

Number = 10x + y = 10(1) + 8 = 18.

Hence, option 2 is the correct option.

Let x be Bharat's present age and y be Rajat's present age,

Given,

Five years ago, Bharat was thrice as old as Rajat.

⇒ x - 5 = 3(y - 5)

⇒ x - 5 = 3y - 15

⇒ x = 3y - 15 + 5

⇒ x = 3y - 10     ....(1)

Given,

Ten years later, Bharat will be twice as old as Rajat,

⇒ x + 10 = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x = 2y + 20 - 10

⇒ x = 2y + 10     ....(2)

Substituting value of x from equation (1) in x = 2y + 10, we get :

⇒ 3y - 10 = 2y + 10

⇒ 3y - 2y = 10 + 10

⇒ y = 20 years.

Substituting value of y in equation (1), we get :

⇒ x = 3y - 10

⇒ x = 3(20) - 10

⇒ x = 60 - 10

⇒ x = 50 years.

The difference between their present ages = x - y = 50 - 20 = 30 years.

Hence, option 3 is the correct option.

Let x be the number of one rupee coins and y be the number of 50 paisa coins in the bag initially.

Given,

Initial total amount = ₹ 140.

⇒ x + 0.5y = 140

⇒ x = 140 - 0.5y     .......(1)

Given,

After replacing half of the 1-rupee coins with 50-paisa coins, the amount becomes ₹ 115.

⇒ $\frac{x}{2} \times 1$ + 0.5y + 0.5 $\Big(\frac{x}{2}\Big)$ = 115

⇒ 0.5x + 0.5y + 0.25x = 115

⇒ 0.75x + 0.5y = 115     ........(2)

Substituting value of x from equation (1) in (2), we get :

⇒ 0.75(140 - 0.5y) + 0.5y = 115

⇒ 105 - 0.375y + 0.5y = 115

⇒ 0.125y = 115 - 105

⇒ 0.125y = 10

⇒ y = $\frac{10}{0.125}$ = 80.

Substituting value of y in equation (1), we get :

⇒ x = 140 - 0.5y

⇒ x = 140 - 0.5(80)

⇒ x = 140 - 40

⇒ x = 100.

∴ Number of one rupee coins = 100 and number of fifty paisa coins = 80.

Hence, option 1 is the correct option.

Let X's speed and Y's speed be x km/hr and y km/hr respectively.

Time = $\frac{\text{Distance}}{\text{Speed}}$

Given,

X takes 3 hours more than Y to walk 30 km.

⇒ $\frac{30}{x} = \frac{30}{y} + 3$

⇒ $\frac{30}{y} = \frac{30}{x} - 3$ .........(1)

Given,

If X doubles his race, he is able to be ahead of Y by $1\frac{1}{2}$ hours.

⇒ $\frac{30}{2x} = \frac{30}{y} - \frac{3}{2}$

⇒ $\frac{30}{y} = \frac{30}{2x} + \frac{3}{2}$ .........(2)

From equation (1) and (2), we get :

$$\begin{aligned}\Rightarrow \frac{30}{x} - 3 = \frac{30}{2x} + \frac{3}{2} \\[1em] \Rightarrow \frac{30}{x} - \frac{30}{2x} = \frac{3}{2} + 3 \\[1em] \Rightarrow \frac{60 - 30}{2x} = \frac{3 + 6}{2} \\[1em] \Rightarrow \frac{30}{2x} = \frac{9}{2} \\[1em] \Rightarrow x = \frac{30}{9} = \frac{10}{3} \text{ km/hr}.\end{aligned}$$

Substituting value of x in equation (1), we get :

$$\begin{aligned}\Rightarrow \frac{30}{y} = \frac{30}{\frac{10}{3}} - 3 \\[1em] \Rightarrow \frac{30}{y} = \frac{90}{10} - 3 \\[1em] \Rightarrow \frac{30}{y} = 9 - 3 \\[1em] \Rightarrow \frac{30}{y} = 6 \\[1em] \Rightarrow y = \frac{30}{6} = 5 \text{ km/hr}.\end{aligned}$$

Hence, option 1 is the correct option.

Let x be the speed of the boat in still water and y be the speed of current,

Downstream speed = (x + y) km/hr

Upstream speed = (x - y) km/hr

Time = $\frac{\text{Distance}}{\text{Speed}}$

Given,

It takes 10 hours to go 44 km downstream and 30 km upstream.

⇒ $\frac{44}{x + y} + \frac{30}{x - y} = 10$ .........(1)

Given,

It takes 13 hours to go 55 km downstream and 40 km upstream.

⇒ $\frac{55}{x + y} + \frac{40}{x - y} = 13$ ........(2)

Substituting $\frac{1}{x + y} = u, \frac{1}{x - y} = v$, in equation (1),

⇒ 44u + 30v = 10     ....(3)

Substituting $\frac{1}{x + y} = u, \frac{1}{x - y} = v$, in equation (2),

⇒ 55u + 40v = 13     ....(4)

Multiply equation (3) by 4, we get :

⇒ 4(44u + 30v = 10)

⇒ 176u + 120v = 40     ....(5)

Multiply equation (4) by 3, we get :

⇒ 3(55u + 40v = 13)

⇒ 165u + 120v = 39     ....(6)

Subtracting equation (5) from equation (6), we get :

⇒ (165u + 120v) - (176u + 120v) = 39 - 40

⇒ (165u + 120v - 176u - 120v) = -1

⇒ -11u = -1

⇒ u = $\frac{-1}{-11} = \frac{1}{11}$.

Substituting value of u in equation (3), we get :

⇒ 44u + 30v = 10

⇒ $44 \times \frac{1}{11}$ + 30v = 10

⇒ 4 + 30v = 10

⇒ 30v = 10 - 4

⇒ 30v = 6

⇒ v = $\frac{6}{30} = \frac{1}{5}$.

$$\begin{aligned}\Rightarrow \frac{1}{x + y} = u \\[1em] \Rightarrow \frac{1}{x + y} = \frac{1}{11} \\[1em] \Rightarrow x + y = 11 \text{ ........(7) } \\[1em] \Rightarrow \frac{1}{x - y} = v \\[1em] \Rightarrow \frac{1}{x - y} = \frac{1}{5} \\[1em] \Rightarrow x - y = 5 \text{ .........(8) }\end{aligned}$$

Adding equations (7) and (8) we get,

⇒ x + y + x - y = 11 + 5

⇒ 2x = 16

⇒ x = $\frac{16}{2} = 8$.

Substituting value of x in equation (8),

⇒ x - y = 5

⇒ 8 - y = 5

⇒ 8 - 5 = y

⇒ y = 3.

The speed of the boat in still water is 8 km/hr and the speed of the current is 3 km/hr.

Hence, option 4 is the correct option.

Let x be the number of boys and y be the number of girls,

Given,

Case 1:

If each boy is given 3 mangoes, then each girl gets 6 mangoes.

⇒ 3x + 6y = 42     .......(1)

Case 2:

If each boy is given 5 mangoes, then each girl gets 3 mangoes.

⇒ 5x + 3y = 42     ....(2)

Multiply equation by 2 we get,

⇒ 2(5x + 3y = 42)

⇒ 10x + 6y = 84     ......(3)

Subtracting equation (1) from (3), we get:

⇒ 10x + 6y - (3x + 6y) = 84 - 42

⇒ 10x + 6y - 3x - 6y = 84 - 42

⇒ 7x = 42

⇒ x = $\frac{42}{7}$ = 6.

Substituting value of x in equation 1, we get :

⇒ 3x + 6y = 42

⇒ 3(6) + 6y = 42

⇒ 18 + 6y = 42

⇒ 6y = 42 - 18

⇒ 6y = 24

⇒ y = $\frac{24}{6}$ = 4.

Thus, no. of boys = 6, no. of girls = 4.

Hence, option 2 is the correct option.

We know that,

Sum of all interior angles of a quadrilateral = 360°.

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ 2x° + 6y° + 10° + 2x° + y° + x° + 10° = 360°

⇒ 5x° + 7y° + 20° = 360°

⇒ 5x° + 7y° = 340°

Substituting x = 40°, y = 20° in L.H.S. of the above equation, we get :

⇒ 5 × 40° + 7 × 20°

⇒ 200° + 140°

⇒ 340°.

Since, L.H.S. = R.H.S.

Solution : x = 40°, y = 20°.

Hence, option 1 is the correct option.

1. Given,

From picture,

x = Length of Bedroom 1, y = Length of kitchen

Length of bathroom = 2

Equations:

Length of rectangular flat = x + y + 2

In rectangle, opposites sides are equal.

⇒ x + y + 2 = 15

⇒ x + y = 15 - 2

⇒ x + y = 13     ....(1)

Area = length × breadth

From figure,

Area of each Bedroom = 5 × x

Area of Kitchen = 5 × y

Given,

Area of two bedrooms and kitchen is 95.

⇒ 2 × (5 × x) + 5 × y = 95

⇒ 10x + 5y = 95

⇒ 5(2x + y) = 95

⇒ 2x + y = $\frac{95}{5}$

⇒ 2x + y = 19

⇒ 2x + y = 19     ....(2)

Subtracting equation (1) from (2), we get :

⇒ 2x + y - (x + y) = 19 - 13

⇒ x = 6 m.

Substituting value of x in equation (1), we get :

⇒ 6 + y = 13

⇒ y = 13 - 6 = 7 m.

Hence, Option (b) is the correct option.

2. From figure,

⇒ Breadth = 15 m

⇒ Length = 5 + 2 + 5 = 12 m

Perimeter = 2(l + b)

= 2(12 + 15)

= 2 × 27

= 54 m.

Hence, Option (a) is the correct option.

3. Given,

Area of bedroom1 = l × b

= x × 5

= 6 × 5

= 30 m².

Area of kitchen = l × b

= y × 5

= 7 × 5

= 35 m².

Area of Bedroom 1 + Kitchen = 30 + 35 = 65 m².

Hence, option (c) is the correct option.

4. From figure,

Area of bedroom 2 = l × b = 5 × x = 5 × 6 = 30 m².

Area of bedroom2 + living room = l × b

= 15 × 7

= 105 m².

Area of living room = 105 - area of bedroom2 = 105 - 30 = 75 m².

Hence, option (d) is the correct option.

5. Given,

Area of kitchen = l × b

= y × 5

= 7 × 5

= 35 m².

Cost of laying tiles on the floor of the kitchen at the rate of ₹ 200 per sq m = 35 × 200 = ₹ 7,000.

Hence, Option (a) is the correct option.

1. Given,

Let x = number of minutes, and y = total cost in ₹

Company P charges: ₹40/month + ₹0.5/min talk time

⇒ y = 0.5x + 40.

Hence, Option (a) is the correct option.

2. Given,

Company Q charges: ₹30/month + ₹1/min talk time

⇒ y = x + 30

Hence, Option (b) is the correct option.

3. If the expenditure is equal, then equating the values of y from part (1) and (2)

⇒ 0.5x + 40 = x + 30

⇒ 40 - 30 = x - 0.5x

⇒ 10 = 0.5x

⇒ x = $\frac{10}{0.5}$ = 20 minutes.

Hence, Option (c) is the correct option.

4. Given,

Manisha used 400 minutes on Company P. Calculating, the money spent by her,

⇒ y = 0.5x + 40

⇒ y = 0.5(400) + 40

⇒ y = 200 + 40

⇒ y = ₹ 240.

Hence, option (a) is the correct option.

5. Given,

Anurag uses 300 minutes,

Cost if he took Company P's plan.

⇒ y = 0.5x + 40

⇒ y = 0.5(300) + 40

⇒ y = 150 + 40

⇒ y = ₹ 190

Cost if he took Company Q's plan.

⇒ y = 1x + 30

⇒ y = 300 + 30

⇒ y = ₹ 330.

Thus, Anurag gets benefit if he uses company P's plan.

Hence, option (a) is the correct option.

1. Tanusha got x, ₹ 100-rupee notes and y, 500-rupee notes and withdraws ₹ 15,000.

⇒ 100x + 500y = 15000

⇒ 100(x + 5y) = 15000

⇒ x + 5y = $\frac{15000}{100}$

⇒ x + 5y = 150.

Hence, Option (a) is the correct option.

2. If she received x number ₹ 100 notes and y number of ₹ 500 notes and total notes are 54.

⇒ x + y = 54

Hence, Option (b) is the correct option.

3. If she withdaws ₹ 16,000 and gets x, ₹ 100 notes and y, 500 notes then.

⇒ 100x + 500y = 16000

⇒ 100(x + 5y) = 16000

⇒ x + 5y = $\frac{16000}{100}$

⇒ x + 5y = 160.

Substituting y = 30 and x = 10 in L.H.S. of the above equation, we get :

⇒ 10 + 5(30) = 10 + 150 = 160 = R.H.S.

Thus, no. of ₹ 100 notes = 10 and no. of ₹ 500 notes = 30.

Hence, Option (d) is the correct option.

4. Given,

She received 25 number ₹ 100 notes and 20 number of ₹ 500 notes,

Amount withdrawn = 100 × 25 + 500 × 20

= 2,500 + 10,000

= ₹ 12,500.

Hence, Option (d) is the correct option.

5. Let x be the number of ₹100 notes and y be the number of ₹500 notes.

The total amount is 100x + 500y.

Can Tanusha withdraw ₹ 10,050.

This means we need to check if the equation 100x + 500y = 10050 has integer solutions for x and y.

The value of x ₹100 notes is 100x, which is a multiple of 100.

The value of y ₹500 notes is 500y, which is also a multiple of 100.

The sum of these two values, the total amount withdrawn, must also be a multiple of 100.

Since, ₹ 10,050 is not a multiple of 100.

Therefore, it is not possible to form the amount ₹ 10,050 using only ₹ 100 and ₹ 500 notes.

Hence, Option (b) is the correct option.

Given,

Equations: 8x + 7y = 37 and 7x + 8y = 38

⇒ 8x + 7y = 37

⇒ 8x = 37 - 7y

⇒ x = $\frac{37 - 7y}{8}$     .....(1)

Substituting value of x from equation (1) in 7x + 8y = 38, we get :

$$\begin{aligned}\Rightarrow 7\Big(\frac{37 - 7y}{8}\Big) + 8y = 38 \\[1em] \Rightarrow \frac{259 - 49y}{8} + 8y = 38 \\[1em] \Rightarrow \frac{259 - 49y + 64y}{8} = 38\\[1em] \Rightarrow 259 + 15y = 38 \times 8 \\[1em] \Rightarrow 259 + 15y = 304 \\[1em] \Rightarrow 15y = 304 - 259 \\[1em] \Rightarrow 15y = 45\\[1em] \Rightarrow y = \frac{45}{15} = 3.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{37 - 7y}{8} \\[1em] \Rightarrow x = \frac{37 - 7 \times 3}{8} \\[1em] \Rightarrow x = \frac{37 - 21}{8} \\[1em] \Rightarrow x = \frac{16}{8} = 2.\end{aligned}$$

x = 2 and y = 3.

∴ Assertion (A) is false.

⇒ ax + by = c and bx + ay = d are simultaneous linear equations in two variables x and y.

∴ Reason (R) is false.

Hence, option 4 is the correct option.

Given,

Equations: $\frac{2}{m} + \frac{3}{m} = 0 \text{ and } \frac{2}{3m} + \frac{2}{n} = \frac{1}{6}$

$$\begin{aligned}\Rightarrow \frac{2}{m} + \frac{3}{m} = 0 \\[1em] \Rightarrow \frac{5}{m} = 0 \\[1em] \Rightarrow 5 = 0 \times m \\[1em] \Rightarrow 5 \ne 0 .\end{aligned}$$

Since 5 is not equal to 0, this equation has no solution for m. We cannot find values for m and n that satisfy both equations simultaneously.

$\Rightarrow \frac{2}{3m} + \frac{2}{n} = \frac{1}{6}$ Again, this is not linear in variables m and n, because the variables are in denominators.

Assertion (A) is false.

An equation of the form ax + by + c = 0, a ≠ 0, b ≠ 0, is called a linear equation in two variables.

Reason(R) is true.

A is false, R is true

Hence, option 2 is the correct option.

A pair of linear equations in two variables cannot have more than one solution. When the pair is consistent and independent, it has exactly one unique solution.

∴ Assertion (A) is true.

If we solve a pair of linear equations in two variables, first by elimination method and then by cross multiplication method, then in some cases the two solutions so obtained may be different.

This is false statement because both methods give same final answers.

∴ Reason(R) is false.

A is true, R is false.

Hence, option 1 is the correct option.

Given,

0.4x + 0.3y = 2.3 and 2.5x - 2y = -5

Solving first equation,

⇒ 0.4x + 0.3y = 2.3

⇒ 10(0.4x + 0.3y = 2.3)     [Multiplying both sides by 10]

⇒ 4x + 3y = 23

⇒ 4x = 23 - 3y

⇒ x = $\frac{23 - 3y}{4}$     ....(1)

⇒ 2.5x - 2y = -5     ....(2)

Substituting value of x from equation (1) in 2.5x - 2y = -5, we get :

$$\begin{aligned}\Rightarrow 2.5 \Big(\frac{23 - 3y}{4}\Big) - 2y = -5 \\[1em] \Rightarrow \frac{57.5 - 7.5y}{4} - 2y = -5 \\[1em] \Rightarrow \frac{57.5 - 7.5y - 8y}{4} = -5 \\[1em] \Rightarrow 57.5 - 15.5y = -5 \times 4 \\[1em] \Rightarrow 57.5 - 15.5y = -20 \\[1em] \Rightarrow -15.5y = -20 - 57.5 \\[1em] \Rightarrow -15.5y = -77.5 \\[1em] \Rightarrow y = \frac{-77.5}{-15.5} = 5.\end{aligned}$$

Substituting value of y in equation (1), we get :

$$\begin{aligned}\Rightarrow x = \frac{23 - 3y}{4} \\[1em] \Rightarrow x = \frac{23 - 3 \times 5}{4} \\[1em] \Rightarrow x = \frac{23 - 15}{4} \\[1em] \Rightarrow x = \frac{8}{4} = 2.\end{aligned}$$

x = 2 and y = 5.

xy = 10.

Hence, option 1 is the correct option.

Given,

Equations: ax - by = a² + b² and $\frac{x + y}{2} = a$

Solving equation $\frac{x + y}{2} = a$,

$$\begin{aligned}\Rightarrow \frac{x + y}{2} = a \\[1em] \Rightarrow x + y = a \times 2 \\[1em] \Rightarrow x + y = 2a \\[1em] \Rightarrow y = 2a - x \text{ ......(1) }\end{aligned}$$

⇒ ax - by = a² + b²     .......(2)

Substituting value of y from equation (1) in (2) we get,

⇒ ax - by = a² + b²

⇒ ax - b(2a - x) = a² + b²

⇒ ax - 2ab + bx = a² + b²

⇒ ax + bx = a² + b² + 2ab

⇒ x(a + b) = a² + b² + 2ab

⇒ x(a + b) = (a + b)²

⇒ x = $\frac{(a + b)^2}{a + b}$

⇒ x = (a + b).

Substituting value of x in equation (1),

⇒ y = 2a - x

⇒ y = 2a - (a + b)

⇒ y = 2a - a - b

⇒ y = a - b

Now,

⇒ x - y = (a + b) - (a - b)

⇒ x - y = (a + b - a + b)

⇒ x - y = 2b.

Hence, option 4 is the correct option.

Given,

Equations:

⇒ 8x + 9y = 42xy

⇒ 2x + 3y = 12xy

Dividing both the sides of first equation by xy, we get :

$$\begin{aligned}\Rightarrow \frac{8x + 9y}{xy} = \frac{42xy}{xy} \\[1em] \Rightarrow \frac{8x}{xy} + \frac{9y}{xy} = 42 \\[1em] \Rightarrow \frac{8}{y} + \frac{9}{x} = 42 \text{ .......(1)}\end{aligned}$$

Dividing both the sides of second equation by xy, we get :

$$\begin{aligned}\Rightarrow \frac{2x + 3y}{xy} = \frac{12xy}{xy} \\[1em] \Rightarrow \frac{2x}{xy} + \frac{3y}{xy} = 12 \\[1em] \Rightarrow \frac{2}{y} + \frac{3}{x} = 12.\end{aligned}$$

Multiplying both sides of the above equation by 4, we get :

$$\begin{aligned}\Rightarrow 4\Big(\frac{2}{y} + \frac{3}{x}\Big) = 12 \times 4 \\[1em] \Rightarrow \frac{8}{y} + \frac{12}{x} = 48 \text{ .......(2)}\end{aligned}$$

Subtracting equation (1) from (2), we get:

$$\begin{aligned}\Rightarrow \frac{8}{y} + \frac{12}{x} - \Big(\frac{8}{y} + \frac{9}{x}\Big) = 48 - 42 \\[1em] \Rightarrow \frac{8}{y} + \frac{12}{x} - \frac{8}{y} - \frac{9}{x} = 6 \\[1em] \Rightarrow \frac{12}{x} - \frac{9}{x} = 6 \\[1em] \Rightarrow \frac{12 - 9}{x} = 6 \\[1em] \Rightarrow \frac{3}{x} = 6 \\[1em] \Rightarrow x = \frac{3}{6} = \frac{1}{2}.\end{aligned}$$

Substituting value of x in equation (1), we get:

$$\begin{aligned}\Rightarrow \frac{8}{y} + \frac{9}{x} = 42 \\[1em] \Rightarrow \frac{8}{y} + \frac{9}{\frac{1}{2}} = 42 \\[1em] \Rightarrow \frac{8}{y} + 18 = 42 \\[1em] \Rightarrow \frac{8}{y} = 42 - 18 \\[1em] \Rightarrow \frac{8}{y} = 24 \\[1em] \Rightarrow y = \frac{8}{24} = \frac{1}{3}.\end{aligned}$$

Calculating xy,

xy = $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.

$\therefore \frac{1}{xy} = 6$

Hence, option 3 is the correct option.

Let cost price of the table be ₹ x and cost price of the chair be ₹ y.

According to case 1 :

⇒ Profit on table = 10%

Selling Price of table = Cost price (1 + Profit%) = $x \Big(1 + \frac{10}{100}\Big)$ = ₹ x × 1.10

⇒ Profit on chair = 25%

Selling Price of chair = Cost price (1 + Profit%) = $y \Big(1 + \frac{25}{100}\Big)$ = ₹ y × 1.25

⇒ 1.10x + 1.25y = 1050

Multiply the equation by 100,

⇒ 100(1.10x + 1.25y) = 100 × 1050

⇒ 110x + 125y = 105000

⇒ 5(22x + 25y) = 5 × 21000

⇒ 22x + 25y = 21000

⇒ 22x = 21000 - 25y

⇒ x = $\frac{21000 - 25y}{22}$     ......(1)

According to case 2 :

⇒ Profit on table = 25%

Selling Price of table = Cost price (1 + Profit%) = $x \Big(1 + \frac{25}{100}\Big)$ = ₹ x × 1.25

⇒ Profit on chair = 10%

Selling Price of chair = Cost price (1 + Profit%) = $y \Big(1 + \frac{10}{100}\Big)$ = ₹ y × 1.10

⇒ 1.25x + 1.10y = 1065

Multiply the equation by 100,

⇒ 100(1.25x + 1.10y) = 1065 × 100

⇒ 125x + 110y = 106500

⇒ 5(25x + 22y) = 5 × 21300

⇒ 25x + 22y = 21300     .......(2)

Substituting value of x from equation 1 in (2), we get :

$$\begin{aligned}\Rightarrow 25 \Big(\frac{21000 - 25y}{22}\Big) + 22y = 21300 \\[1em] \Rightarrow \frac{25(21000 - 25y) + 484y}{22} = 21300 \\[1em] \Rightarrow 525000 - 625y + 484y = 468600 \\[1em] \Rightarrow -141y = 468600 - 525000 \\[1em] \Rightarrow -141y = -56400 \\[1em] \Rightarrow y = \frac{-56400}{-141} = 400.\end{aligned}$$

Substituting value of y in equation 1, we get :

$$\begin{aligned}\Rightarrow x = \frac{21000 - 25y}{22} \\[1em] \Rightarrow x = \frac{21000 - 25 \times 400}{22} \\[1em] \Rightarrow x = \frac{21000 - 10000}{22} \\[1em] \Rightarrow x = \frac{11000}{22} = 500.\end{aligned}$$

Hence, cost Price of Table = ₹ 500 and cost Price of Chair = ₹ 400.

Let x be speed of the stream.

Given,

Speed of boat in still water = 5 km/hr.

Speed of boat in upstream = (5 - x) km/hr

Speed of boat in downstream = (5 + x) km/hr

By formula,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Given,

The boatman takes thrice as much time in going 40 km upstream as in going 40 km downstream.

$$\begin{aligned}\therefore \frac{40}{5 - x} = 3 \times \frac{40}{5 + x} \\[1em] \Rightarrow \frac{1}{5 - x} = \frac{1}{5 + x} \\[1em] \Rightarrow 5 + x = 3(5 - x) \\[1em] \Rightarrow 5 + x = 15 - 3x \\[1em] \Rightarrow x + 3x = 15 - 3 \\[1em] \Rightarrow 4x = 10 \\[1em] \Rightarrow x = \frac{10}{4} \\[1em] \Rightarrow x = 2.5 \text{ km/hr }\end{aligned}$$

Hence, the speed of stream = 2.5 km/hr.

Let x be the number of pens sold and y be the number of pencils sold.

Given,

Cost Price (CP) of 1 pen = ₹ 5

Profit on pens = 12%

⇒ SP of 1 pen = CP of pen + (Profit % of CP)

⇒ SP of 1 pen = 5 + $\frac{12}{100} \times 5$ = 5 + 0.12 × 5 = 5 + 0.60 = ₹ 5.60

Given,

Cost Price (CP) of 1 pencil = ₹ 1

Profit on pencils = 10%

SP of 1 pencil = CP of pencil + (Profit % of CP)

SP of 1 pencil = 1 + $\frac{10}{100}$ × 1 = 1 + 0.10 = ₹ 1.10

For every two pens, he buys three pencils,

This means the ratio of pens to pencils is x : y = 2 : 3.

$\frac{x}{y} = \frac{2}{3}$

⇒ y = $\frac{3}{2}x$     .......(1)

Given,

Total sale is ₹ 725,

⇒ x × 5.60 + y × 1.10 = 725

⇒ 5.60x + 1.10y = 725     ......(2)

Substitute the expression for y from equation (1) in (2), we get :

⇒ 5.60x + 1.10 $\times \frac{3}{2}x$ = 725

⇒ 5.60x + 1.10 × 1.5 = 725

⇒ 5.60x + 1.65x = 725

⇒ 7.25x = 725

⇒ y = $\frac{725}{7.25}$ = 100.

Substituting value of y in equation 1 we get,

⇒ y = $\frac{3}{2} \times 100$

⇒ y = 3 × 50

⇒ y = 150.

Hence, the number of pens sold = 100 and number of pencils sold = 150.

Given,

The three angles of the triangle in ascending order are x, y, and z.

Given,

⇒ y − x = 10°

⇒ y = 10° + x     ....(1)

Given,

⇒ z − y = 10°     ....(2)

Substitute value of y from equation (1) in (2), we get :

⇒ z − y = 10°

⇒ z - (10° + x) = 10°

⇒ z = 10° + 10° + x

⇒ z = 20° + x.

So, the three angles of the triangle in terms of x are:

First angle: x

Second angle: x + 10°

Third angle: x + 20°

We know that,

The sum of the angles in any triangle is always 180°.

⇒ x + y + z = 180°

⇒ x + (x + 10°) + (x + 20°) = 180°

⇒ 3x + 30° = 180°

⇒ 3x = 180° − 30°

⇒ 3x = 150°

⇒ x = $\frac{150°}{3}$

⇒ x = 50°.

Substituting the value of x,

⇒ y = x + 10° = 50° + 10° = 60°

⇒ z = x + 20° = 50° + 20° = 70°.

Hence, x = 50°, y = 60°, z = 70°.